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Introduction to algebra

Table of Contents

Introduction

Welcome to the Introduction to algebra module. This module gives an overview of key ideas in algebra. These include how to rearrange and solve simple linear and quadratic equations, how to work with exponents and logarithms, and how to recognise and define functions.

You can work through the whole module in the order it is presented. You can also choose to skip to specific pages or sections, depending on your needs.

Your feedback on this module is very welcome. You can provide feedback at any time using the feedback page. If you have any questions about the module, please contact Library-UniSkills@curtin.edu.au

What you will learn

To solve algebraic equations and find unknown values, which is a key part of algebra, you first need to understand the main terms and skills explained on this page.

To solve algebraic equations and find unknown values, which is a key part of algebra, you first need to understand the main terms and skills explained on this page.

In brief, you will learn:

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Definitions of important terms

When a problem is written using algebra it can be called either:

Both algebraic expressions and algebraic equations involve the use of constants and variables:

Constants are numbers wthat have a fixed value (for example \(1, 5, 2108, 37.5, -103,\) etc.).

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Variables are usually small letters of the alphabet, or sometimes other symbols. They are used to represent an unknown number in an expression or equation (for example \(x, y, a, \theta,\) etc.). The value of this number can change (or vary, as the word variable suggests) depending on the situation.

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In the algebraic expression \(a + 5,\) for example, \(5\) is a constant, while \(a\) is a variable. This means that different numbers can be substituted for \(a\) and the value of the expression will change accordingly. For example, if \(a = 1\) then \(a + 5\) becomes \(1 + 5,\) which we know is equal to \(6\).

Another important idea is that each ‘part’ of an algebraic expression or equation separated is called a term. A term is separated from other terms by addition, subtraction, or division. A term can be a constant, a variable, or a combination of both.

For example, the following algebraic expression contains \(5\) different terms, each separated by addition (note that we usually write \(x\) and \(y\) rather than \(1x\) and \(1y\), even though they mean the same thing):

\[2x + y + 3y + 5x + x\]

Because this expression has more than one variable, it includes both like terms and unlike terms.

Like terms are terms which have the same variable. For example \(2x,\) \(5x\) and \(x\) are all like terms in the expression above. Similarly, \(y\) and \(3y\) are also like terms.

Note that \(x\) and \(x^2\) are not like terms because one term is \(x\) while the other has \(x\) raised to the power of \(2.\) In the same way, \(x\) and \(xy\) are not like terms because one is \(x\) only while the other has \(x\) and \(y.\)

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Unlike terms are terms that have different variables. For example \(2x\) and \(y\) are unlike terms. As shown above, \(x\) and \(x^2\) are also unlike terms, and so are \(x\) and \(xy\).

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It is important to understand that only like terms can be added or subtracted. For example, you would not say that \(2\) apples plus \(5\) oranges equals \(7\) apples (or \(7\) oranges), In the same way, you cannot add \(2x\) and \(5y\) together.

Finally, when terms are combined in certain ways, they form polynomials.

A polynomial is an expression made up of constants and/or variables raised to positive powers (also known as exponents), combined using addition, subtraction, and/or multiplication. For example, \(x^2 + 2x - 3\) is a polynomial, and so is \(20\), but \(x^{-2} + 3\) is not.

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Simplifying algebraic expressions

As with most topics, it is best to write an algebraic expression or equation in the simplest form possible. There is no need to make it more complicated than necessary. There are a few key methods you can use to simplify algebraic expressions and equations, and these are explained in this section.

Note that the examples focus on algebraic expressions, as these are easier to use when first learning the techniques. However, remember that the main goal is to apply these techniques to solve algebraic equations, as shown in later sections of this module.

Adding or subtracting like terms

You can simplify an algebraic expression by adding or subtracting like terms. To do this, follow these steps:

  1. Rewrite the expression so that all like terms are grouped together. When moving terms, make sure you move the \(+\) or \(–\) sign in front of each term with it.

  2. Simplify the expression by adding or subtracting like terms.

Example: Simplify the algebraic expression \(2a\,–\,3b\,–\,a\,+\,4ab\,–\,2b\)

Solution: Following the steps above, we have:

  1. Rewriting the expression so that all like terms are next to each other gives \(2a\,–\,a\,–\,3b\,–\,2b\,+\,4ab\) (or you can write it in other ways, as long as all the \(a\) terms are together and all the \(b\) terms are together; the \(ab\) term doesn’t have any like terms)

  2. Adding and subtracting the like terms in the rearranged algebraic expression \(2a\,–\,a\,–\,3b\,–\,2b\,+\,4ab\) gives \(2a\,–\,a = a\) and \(-3b\,–\,2b = -5b\), so our simplified expression is \(a\,-\,5b\,+\,4ab\)

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Multiplying or dividing terms

Note that there are several ways to show that two or more terms need to be multiplied. For example, if \(x\) and \(y\) are to be multiplied, this might be written as \(x \times y\), as \(x.y\), as \((x)(y)\) or \(x(y)\).

No matter how the multiplication is written, you can simplify an algebraic expression by multiplying or dividing terms. Follow these steps:

  1. Rewrite the expression (or each part of the expression, when dividing) as a product of its factors.

  2. If needed, rewrite the expression (or each part of the expression, when dividing) so that like terms are grouped together.

  3. Multiply or divide the constants and multiply or divide the variables, as needed. Remember that if you multiply or divide a negative by a negative, the result is positive (for example, \(-2 \times -2 = 4\)).

Example 1: Simplify the algebraic expression \(4ab \times 3acd\)

Solution: Following the steps above, we have:

  1. Rewriting the expression as a product of its factors gives \(4 \times a \times b \times 3 \times a \times c \times d\)

  2. Rewriting the expression so that the like terms are next to each other gives \(4 \times 3 \times a \times a \times b \times c \times d\)

  3. Since \(4 \times 3 = 12\) and \(a \times a = a^2\), it follows that the simplified expression is \(12a^2bcd\)

Note that \(a^2\) is an example of an exponent (otherwise known as an indice or power). If you are unsure of how to work with exponents, it is recommended that you visit the Exponents page of this module before proceeding with further examples on this page.

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Example 2: Simplify the algebraic expression \(\frac{20x^3y}{4x^2y^2}\)

Solution: Following the steps above, we have:

  1. Rewriting each part of the expression as a product of its factors gives \(\frac{20 \times x \times x \times x \times y}{4 \times x \times x \times y \times y}\)

  2. The like terms are already next to each other for both parts of this expression, so rewriting is not required

  3. Since \(\frac{20}{4} = 5\) and \(\frac{x \times x \times x}{x \times x} = x\) and \(\frac{y}{y \times y} = \frac{1}{y}\), it follows that the simplified expression is \(\frac{5x}{y}\)

Note that the above division above has been performed by simplifying fractions, which works in the same way for variables as it does for constants. If you would like a refresher on how simplifying fractions works, please visit the Fractions section of the Numeracy fundamentals module.

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After you have worked through the examples above, try simplifying some or all of the following algebraic expressions (again, visit the Exponents page of this module first if needed).

1. \(c + 5c - 2c\)

ANSWER: This expression contains all like terms, which can be added and subtracted to give \(4c\)

2. \(3d – 5c + 4 – d – 9 + 3c\)

ANSWER: Rearranging this expression gives \(3d - d - 5c + 3c + 4 - 9\), which can then be simplified to \(2d - 2c -5\)

3. \((13a)(2b)\)

ANSWER: Rewriting this expression gives \(13 \times a \times 2 \times b\), which then gives \(13 \times 2 \times a \times b\) and therefore simplifies to \(26ab\)

4. \(\frac{10a^2b}{5ac}\)

ANSWER: Rewriting this expression gives \(\frac{10 \times a \times a \times b}{5 \times a \times c}\), and since \(\frac{10}{5} = 2\) and \(\frac{a \times a}{a} = a\), it follows that it simplifies to \(\frac{2ab}{c}\)

5. \(3a + b – a – 3b\)

ANSWER: Rearranging this expression gives \(3a - a + b - 3b\), which can then be simplified to \(2a - 2b\)

6. \(2f + 3h – g + 5h + g\)

ANSWER: Rearranging this expression gives \(2f + 3h + 5h - g + g\), which can then be simplified to \(2f + 8h\)

7. \(20c – 4d + 5c + 10 – 15d\)

ANSWER: Rearranging this expression gives \(20c + 5c - 4d - 15d + 10\), which can then be simplified to \(25c - 19d + 10\)

8. \(-5f + 6 + 102g + 10f + 3\)

ANSWER: Rearranging this expression gives \(-5f + 10f + 102g + 6 + 3\), which can then be simplified to \(5f + 102g + 9\)

9. \(4c \times ab \times 9\)

ANSWER: Rewriting this expression gives \(4 \times c \times a \times b \times 9\), which then gives \(4 \times 9 \times c \times a \times b\) and therefore simplifies to \(36abc\) (while the variables can be listed in any order, the usual convention is to list them alphabetically)

10. \(2a \times a \times 5a^2\)

ANSWER: Rewriting this expression gives \(2 \times a \times a \times 5 \times a \times a\), which then gives \(2 \times 5 \times a \times a \times a \times a\) and therefore simplifies to \(10a^4\)

11. \(\frac{12ac}{ab}\)

ANSWER: Rewriting this expression gives \(\frac{12 \times a \times c}{a \times b}\), and since \(\frac{a}{a} = 1\), it follows that it simplifies to \(\frac{12c}{b}\)

12. \(\frac{20d^2ef}{5de}\)

ANSWER: Rewriting this expression gives \(\frac{20 \times d \times d \times e \times f}{5 \times d \times e}\), and since \(\frac{20}{5} = 4\) and \(\frac{d \times d}{d} = d\) and \(\frac{e}{e} = 1\), it follows that it simplifies to \(4df\)

Expanding brackets

Algebraic expressions or equations often include brackets. When this happens, it can be helpful to expand the brackets to make the expression or equation easier to work with. This section explains how to expand brackets for both one and two sets of brackets. Note again that the examples use algebraic expressions for simplicity. You will learn how to apply these methods to solve algebraic equations in later sections of this module.

Expanding one set of brackets

An algebraic expression that has one term outside a set of brackets can be expanded using the following steps:

  1. Write out each multiplication that needs to be done. Do this by multiplying the term outside the brackets by each term inside the brackets. If you place brackets around each product, it can help you keep track of the steps and deal with negative signs later.

  2. Carry out each multiplication using the method explained earlier. Keep each result in brackets so that it is easier to track any negative signs.

  3. Remove the brackets from each term. If there is a negative sign inside a set of brackets, change the sign of the terms when you remove the brackets.

Example: Expand the brackets in the algebraic expression \(-3a(2b + 4a - 9)\)

Solution: Following the steps above, we have:

  1. Writing out the multiplications we need to perform, using brackets to indicate multiplication and adding each pair of terms, gives \((-3a)(2b) + (-3a)(4a) + (-3a)(-9)\)

  2. Performing the multiplication above, keeping the results in brackets, gives \((-6ab) + (-12a^2) + (27a)\)

  3. Removing brackets and making the required terms negative gives \(-6ab - 12a^2 + 27a\)

Note that \(a^2\) is an example of an exponent (otherwise known as an indice or power). If you are unsure of how to work with exponents, it is recommended that you visit the Exponents page of this module before proceeding with further examples on this page.

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Expanding pairs of brackets

An algebraic expression with two sets of brackets being multiplied can be expanded using the following steps:

  1. Write out each multiplication that needs to be done. Multiply each term in the first set of brackets by each term in the second set of brackets, and then add the results together. Use brackets to show each multiplication and to help avoid mistakes with negative signs. A helpful way to remember which terms to multiply when there are two terms in each set of brackets is with the acronym FOIL:
    • First: Multiply the first terms in each bracket
    • Outside: Multiply the two outside terms
    • Inside: Multiply the two inside terms
    • Last: Multiply the last terms in each bracket
      \(\)

  2. Perform each multiplication, keeping each result in brackets so you can clearly track any negative signs.

  3. Remove the brackets from each term. If there is a negative sign in a set of brackets, change the sign of the terms when you remove the brackets.

  4. Simplify the expression by adding or subtracting like terms, as explained in the earlier section.

Example: Expand the brackets in the algebraic expression \((a - 3)(a + 4)\)

Solution: Following the steps above, we have:

  1. Writing out the multiplications we need to perform, using brackets to indicate multiplication and adding each pair of terms, gives \((a)(a) + (a)(4) + (-3)(a) + (-3)(4)\)
    Again note that FOIL can help with this:

  2. Performing the multiplication above, keeping the results in brackets, gives \((a^2) + (4a) + (-3a) + (-12)\)

  3. Removing brackets and making the required terms negative gives \(a^2 + 4a - 3a - 12\)

  4. Finally, simplifying the expression gives \(a^2 + a - 12\)

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After you have worked through the examples above, try expanding the brackets in some or all of the following algebraic expressions (again, visit the Exponents page of this module first if needed).

1. \(b(3 + b)\)

\(= (b)(3) + (b)(b)\)

\(= 3b + b^2\)

2. \(3xy(6x - y^2 + x^2y)\)

\(= (3xy)(6x) + (3xy)(-y^2) + (3xy)(x^2y)\)

\(= 18x^2y) + (-3xy^3) + (3x^3y^2)\)

\(= 18x^2y - 3xy^3 + 3x^3y^2\)

3. \((e + f)(g + f)\)

\(= (e)(g) + (e)(f) + (f)(g) + (f)(f)\)

\(= eg + ef + fg + f^2\)

4. \((c - 4d)(-c - 3d)\)

\(= (c)(-c) + (c)(-3d) + (-4d)(-c) + (-4d)(-3d)\)

\(= (-c^2) + (-3cd) + (4cd) + (12d^2)\)

\(= -c^2 - 3cd + 4cd + 12d^2\)

\(= -c^2 + cd + 12d^2\)

5. \(a(-a +1 – ac^2) \)

\(= (a)(-a) + (a)(1) + (a)(-ac^2)\)

\(= (-a^2) + (a) + (-a^2c^2)\)

\(= -a^2 + a - a^2c^2\)

6. \(–3d(3 – 2c + 5d)\)

\(= (-3d)(3) + (-3d)(-2c) + (-3d)(5d)\)

\(= (-9d) + (6cd) + (-15d^2)\)

\(= -9d + 6cd - 15d^2\)

7. \(–a^3(-a^2 – 3a + 4)\)

\(= (-a^3)(-a^2) + (-a^3)(-3a) + (-a^3)(4)\)

\(= (a^5) + (3a^4) + (-4a^3)\)

\(= a^5 + 3a^4 - 4a^3\)

8. \(4dg^2(2d + 3g^3)\)

\(= (4dg^2)(2d) + (4dg^2)(3g^3)\)

\(= 8d^2g^2 + 12dg^5\)

9. \((7x + 2y)(8x – 3y)\)

\(= (7x)(8x) + (7x)(-3y) + (2y)(8x) + (2y)(-3y)\)

\(= (56x^2) + (-21xy) + (16xy) + (-6y^2)\)

\(= 56x^2 - 21xy + 16xy - 6y^2\)

\(= 56x^2 - 5xy - 6y^2\)

10. \((x^2 + y)(x + y)\)

\(=(x^2)(x) + (x^2)(y) + (y)(x) + (y)(y)\)

\(= x^3 + x^2y + xy + y^2\)

11. \((x + 3y)(2x – 4)\)

\(= (x)(2x) + (x)(-4) + (3y)(2x) + (3y)(-4)\)

\(= (2x^2) + (-4x) + (6xy) + (-12y)\)

\(= 2x^2 - 4x + 6xy - 12y\)

12. \((2x + 3)(2x – 1)\)

\(= (2x)(2x) + (2x)(-1) + (3)(2x) + (3)(-1)\)

\(= (4x^2) + (-2x) + (6x) + (-3)\)

\(= 4x^2 - 2x + 6x - 3\)

\(= 4x^2 + 4x - 3\)

Factorising

Factorising is the opposite of expanding brackets. Instead of removing brackets, you add them into an algebraic expression or equation. Factorising is another method that can help you solve algebraic equations in some situations (however, it would not be helpful to keep switching between expanding and factorising the same equation).

This section explains how to factorise using one set of brackets, and also looks at a common example of factorising with two sets of brackets. Note that the examples use algebraic expressions for simplicity. You will learn how to apply these methods to solve algebraic equations in later sections of this module.

Factorising with one set of brackets

You can factorise an algebraic expression using one set of brackets by following these steps:

  1. Identify the highest common factor (HCF) of all the terms in the expression. The highest common factor is the largest value that divides into each term. It may include a constant, a variable (or variables), or a combination of both.

  2. Place the highest common factor outside of a set of brackets.

  3. Write the remaining part of each term inside the brackets.

Example: Factorise the algebraic expression \(3x^2 + 21x\)

Solution: Following the steps above, we have:

  1. The highest common factor is \(3x\), since \(3\) divides into each term in the expression a whole number of times and there is one \(x\) in each term.

  2. Putting this outside a set of brackets gives \(3x(…..)\)

  3. Since there is an \(x\) remaining in \(3x^2\) when it is divided through by \(3x\), the first term inside the brackets should be \(x\).
    Since there is a \(7\) remaining in \(21x\) when it is divided through by \(3x\), the second term inside the brackets should be \(7\).
    Therefore it follows that the factorised form of the expression is \(3x(x + 7)\)

Note that \(x^2\) is an example of an exponent (otherwise known as an indice or power). If you are unsure of how to work with exponents, it is recommended that you visit the Exponents page of this module before proceeding with further examples on this page.

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Factorising with two sets of brackets

Sometimes you need two sets of brackets to factorise an expression. A common example is a quadratic expression, which has the form \(ax^2 + bx + c\) (where \(a\), \(b\) and \(c\) are numbers, and \(a\) does not equal 0 - as this would not form a quadratic).

For more information on quadratic equations see the Solving quadratic equations page of this module. For now, you should know that you can factorise a quadratic expression where \(a = 1\) by following these steps:

  1. Write two sets of brackets, and place the variable from the expression inside both brackets.

  2. Find pairs of factors for the last term in the expression. Make sure you consider whether the factors are positive or negative.

  3. Decide which pair of factors can be added or subtracted to give the constant in the middle term of the expression. Then place one factor in each set of brackets, using the correct signs.

Example: Factorise the algebraic expression \(x^2 + 7x + 12\)

Solution: Following the steps above, we have:

  1. Writing out two sets of brackets with the variable inside each gives \((x\textrm{…..})(x\textrm{…..})\)

  2. The pairs of factors for \(12\) are \(1\) and \(12\), \(2\) and \(6\) and \(3\) and \(4\). Note that both terms must be positive, since they need to multiply together to make positive \(12\) and add together to make positive \(7\)

  3. The pair of factors that can be added together to make \(7\) is \(3\) and \(4\), so the factorised form of \(x^2 + 7x + 12\) is \((x + 3)(x + 4)\), or \((x + 4)(x + 3)\) (it doesn’t matter which way around you write it)

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After you have worked through the examples above, try factorising some or all of the following algebraic expressions (again, visit the Exponents page of this module first if needed).

1. \(14a + 8ab\)

Common factor is \(2a\), so factorising gives \(2a(7 + 4b)\)

2. \(12x^3y – 3x^2y^2 + 6x^2yz\)

Common factor is \(3x^2y\), so factorising gives \(3x^2y(4x - y + 2z)\)

3. \(x^2 + x – 12\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-12\) and add together to make \(1\) (since \(x = 1x\)). The factors of \(12\) are \(1\) & \(12\), \(2\) & \(6\) and \(3\) & \(4\), and one of each pair must be negative and one positive in order to multiply together to make \(-12\). The correct pair is \(-3\) and \(4\), since \(-3 + 4 = 1\), so the expression can be factorised as \((x - 3)(x + 4)\)

4. \( y^2 – 8y + 12\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(12\) and add together to make \(-8\). The factors of \(12\) are \(1\) & \(12\), \(2\) & \(6\) and \(3\) & \(4\), and both numbers must be negative in order to multiply together to make \(12\) and add together to make \(-8\). The correct pair is \(-2\) and \(-6\), since \(-2 - 6 = -8\), so the expression can be factorised as \((y - 2)(y - 6)\)

5. \(z^2 – 4z – 12\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-12\) and add together to make \(-4\). The factors of \(12\) are \(1\) & \(12\), \(2\) & \(6\) and \(3\) & \(4\), and one of each pair must be negative and one positive in order to multiply together to make \(-12\). The correct pair is \(2\) and \(-6\), since \(2 - 6 = -4\), so the expression can be factorised as \((z + 2)(z - 6)\)

6. \(10c + 15c^2\)

Common factor is \(5c\), so factorising gives \(5c(2 + 3c)\)

7. \(16abc + 4ac – 12abc^2\)

Common factor is \(4ac\), so factorising gives \(4ac(4b + 1 - 3bc)\)

8. \(3mn + 6m^2 – 12mn^2\)

Common factor is \(3m\), so factorising gives \(3m(n + 2m - 4n^2)\)

9. \(-5x^2y – 10x^3 – 20x^2\)

Common factor is \(-5x^2\), so factorising gives \(-5x^2(y + 2x + 4)\)

10. \(x^2 + 5x + 6\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(6\) and add together to make \(5\). The factors of \(6\) are \(1\) & \(6\) and \(2\) & \(3\), and both numbers must be positive in order to multiply together to make \(6\) and add together to make \(5\). The correct pair is \(2\) and \(3\), since \(2 + 3 = 5\), so the expression can be factorised as \((x + 2)(x + 3)\)

11. \(y^2 - 3y – 10\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-10\) and add together to make \(-3\). The factors of \(10\) are \(1\) & \(10\) and \(2\) & \(5\), and one of each pair must be negative and one positive in order to multiply together to make \(-10\). The correct pair is \(2\) and \(-5\), since \(2 - 5 = -3\), so the expression can be factorised as \((y + 2)(y - 5)\)

12. \(a^2 - 10a + 24\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(24\) and add together to make \(-10\). The factors of \(24\) are \(1\) & \(24\), \(2\) & \(12\), \(3\) & \(8\) and \(4\) & \(6\), and both numbers must be negative in order to multiply together to make \(24\) and add together to make \(-10\). The correct pair is \(-6\) and \(-4\), since \(-6 - 4 = -10\), so the expression can be factorised as \((a - 6)(a - 4)\)

13. \(b^2 - b – 20\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-20\) and add together to make \(-1\). The factors of \(20\) are \(1\) & \(20\), \(2\) & \(10\) and \(4\) & \(5\), and one of each pair must be negative and one positive in order to multiply together to make \(-20\). The correct pair is \(4\) and \(-5\), since \(4 - 5 = -1\), so the expression can be factorised as \((b + 4)(b - 5)\)

14. \(c^2 + 2c + 1\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(1\) and add together to make \(2\). The only factors of \(1\) are \(1\) & \(1\), and we need both of these to be positive in order to multiply together to make \(1\) and add together to make \(2\). Hence the expression can be factorised as \((c + 1)(c + 1)\), which we would usually write as \((c + 1)^2\)

15. \(d^2 + d – 30\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-30\) and add together to make \(1\). The factors of \(30\) are \(1\) & \(30\), \(2\) & \(15\), \(3\) & \(10\) and \(5\) & \(6\), and one of each pair must be negative and one positive in order to multiply together to make \(-30\). The correct pair is \(6\) and \(-5\), since \(6 - 5 = 1\), so the expression can be factorised as \((d + 6)(d - 5)\)

Linear equations

Once you have an understanding of some key algebraic terms and techniques you can apply them to solve linear equations, as detailed in this page.

In brief, it covers the following:

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What is a linear equation?

Recall that an algebraic equation is similar to an algebraic expression, but it contains an equals (\(=\)) sign. For example, \(5 + d\) is an algebraic expression, while \(5 + d = 10\) is an algebraic equation.

A particular kind of algebraic equation is a linear equation, or first degree equation:

A linear equation is an equation where the variables are only raised to the first power. For example \(a\,–\,3 = 5\) is a linear equation, while \(a^2\,–\,3 = 5\) is not.

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A linear equation with only one variable is called a linear equation in one variable. A linear equation with two variables is called a linear equation in two variables, and this pattern continues for more variables.

When you have a linear equation in one variable, you can find the value of that variable by solving the equation. To do this, you rearrange the equation so that the variable is by itself on one side of the equals sign. For example, to solve the equation \(c + 2 = 5\) you need to get \(c\) on its own on one side of the equals sign.

You can also solve linear equations that have two or more variables. However, to do this, you need the same number of equations as variables. This process is called solving simultaneous equations. This topic is not covered here, but if you would like to learn more,one example of a page you might find helpful is https://www.mathsisfun.com/algebra/systems-linear-equations.html.

Rearranging linear equations

To solve a linear equation in one variable, you need to rearrange the equation. This means you remove the constants that are attached to the variable. These constants may be added to the variable or multiplied by it.

To remove these constants, you must undo what has been done to the variable by using the opposite operation. For example you might:

Remove a constant that was… By…
Added Subtracting it
Subtracted Adding it
Multiplied by Dividing by it
Divided by Multiplying by it

The important rule to remember is that any operation you perform on one side of the equation must also be performed on the other side. This keeps the equation balanced.

In other words, whatever you do to one side of an equation, you must also do to the other side in the same step. For example:

On the left hand side (LHS) of \(=\), if you… On the right hand side (RHS) of \(=\), you must…
Add \(2\) (\(+2\)) Add \(2\) (\(+2\))
Subtract \(5\) (\(-5\)) Subtract \(5\) (\(-5\))
Multiply by \(7\) (\(\times 7\)) Multiply by \(7\) (\(\times 7\))
Divide by \(3\) (\(\div 3\)) Divide by \(3\) (\(\div 3\))

Returning to the equation \(c + 2 = 5\), to solve it you need to undo the addition of \(2\). You can do this by subtracting \(2\) from both sides of the equation, as shown below.

Note that for a simple equation like this, you may not need to use these steps. However, these methods are very useful when working with more complex equations. It is best to start with a simple example to understand the process.

\[\begin{aligned} && c + 2 &= 5 \\\ \therefore && c + 2 - 2 &= 5 - 2 && \textrm{ (undo addition by subtracting) } \\\ \therefore && c &= 3 && \textrm{ (simplify - equation is solved!) } \end{aligned}\]

Note that the \(\therefore\) sign shown above means ‘therefore’.

At this stage, it is important to understand that as you become more confident in solving equations, you may develop different ways of showing your working that make sense to you. These methods may look a little different from the example above, and that is okay, as long as your working is correct.

For example, you might think of the ‘plus \(2\)’ as moving to the other side of the equation and changing to ‘subtract \(2\)’, as shown below:

\[\begin{aligned} && c + 2 &= 5 \\\ \therefore && c &= 5 - 2 && \textrm{ (addition of }2\textrm{ on left becomes subtraction of }2\textrm{ on right)} &\\\ \therefore && c &= 3 && \textrm{ (simplify - equation is solved!)} \end{aligned}\]

Steps for solving linear equations

Often, you need to undo more than one operation to solve a linear equation in one variable. This can take some careful thinking for more complex equations. As mentioned earlier, there can be different ways to solve the same equation. However, the following steps are a useful starting point:

  1. If your equation contains brackets, expand them first (if you need help wit this, refer to the Key terms and techniques page of this module).

  2. If your equation contains like terms, group them together next. You may need to add or subtract terms from one side of the equation. Then simplify the equation (again, refer to the Key terms and techniques if needed).

  3. If you need to remove more than one constant from the variable, undo one operation at a time in this order:
    • Remove anything that has been added to or subtracted from the variable
    • Remove anything that the variable has been multiplied or divided by
  4. Check that your answer is correct. To do this, substitute the value you found back into the original equation and evaluate it. If the answer is not correct, go back, find where you made a mistake, and work through the steps again.

Example: Solve the linear equation \(10d - 2 = 7(d + 1)\)

Solution: Following the steps above, we have:

\[\begin{aligned} && 10d - 2 &= 7(d + 1) \\\ \therefore && 10d - 2 &= 7d + 7 && \textrm{ (expand brackets)} \\\ \therefore && 10d - 2 - 7d &= 7d + 7 - 7d && \textrm{ (remove variable from RHS of equation by subtracting)} \\\ \therefore && 3d - 2 &= 7 && \textrm{ (simplify)} \\\ \therefore && 3d - 2 + 2 &= 7 + 2 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 3d &= 9 && \textrm{ (simplify)} \\\ \therefore && 3d/3 &= 9/3 && \textrm{ (undo multiplication by dividing)} \\\ \therefore && d &= 3 && \textrm{ (simplify - equation is solved!)} \end{aligned}\]

Finally, substituting \(3\) back into the original equation gives \((10)(3) - 2 = 30 - 2 = 28\) on the left hand side, and \(7(3 + 1) = 7(4) = 28\) on the right hand side. So the solution is correct.

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Once you have worked through the example above, have a go at solving some or all of the following linear equations:

1. Solve \(c + 3 = 9\) to find the value of \(c\)

\(\begin{aligned} && c + 3 &= 9 \\\ \therefore && c + 3 - 3 &= 9 - 3 && \textrm{ (undo addition by subtracting)} \\\ \therefore && c &= 6 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

2. Solve \(\frac{a}{3} = 2\) to find the value of \(a\)

\(\begin{aligned} && \frac{a}{3} &= 2 \\\ \therefore && 3(\frac{a}{3}) &= 3(2) && \textrm{ (undo division by multiplying)} \\\ \therefore && a &= 6 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

3. Solve \(3p + 2 = 17\) to find the value of \(p\)

\(\begin{aligned} && 3p + 2 &= 17 \\\ \therefore && 3p + 2 - 2 &= 17 - 2 && \textrm{ (undo addition by subtracting)} \\\ \therefore && 3p &= 15 && \textrm{ (simplify)} \\\ \therefore && \frac{3p}{3} &= \frac{15}{3} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && p &= 5 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

4. Solve \(3y - 5 = 13\) to find the value of \(y\)

\(\begin{aligned} && 3y - 5 &= 13 \\\ \therefore && 3y - 5 + 5 &= 13 + 5 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 3y &= 18 && \textrm{ (simplify)}& \\\ \therefore && \frac{3y}{3} &= \frac{18}{3} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && p &= 6 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

5. Solve \(3a + 4 = a + 6\) to find the value of \(a\)

\(\begin{aligned} && 3a + 4 &= a + 6 \\\ \therefore && 3a + 4 - a &= a + 6 - a && \textrm{ (remove variable from RHS by subtracting)} \\\ \therefore && 2a + 4 &= 6 && \textrm{ (simplify)} \\\ \therefore && 2a + 4 - 4 &= 6 - 4 && \textrm{ (undo addition by subtracting)} \\\ \therefore && 2a &= 2 && \textrm{ (simplify)} \\\ \therefore && \frac{2a}{2} &= \frac{2}{2} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && a &= 1 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

6. Solve \(6n = 20 + 2n\) to find the value of \(n\)

\(\begin{aligned} && 6n &= 20 + 2n \\\ \therefore && 6n - 2n &= 20 + 2n - 2n && \textrm{ (remove variable from RHS by subtracting)} \\\ \therefore && 4n &= 20 && \textrm{ (simplify)} \\\ \therefore && \frac{4n}{4} &= \frac{20}{4} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && n &= 5 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

7. Solve \(5a – 5 + 10a = -5\) to find the value of \(a\)

\(\begin{aligned} && 5a - 5 + 10a &= -5 \\\ \therefore && 15a - 5 &= -5 && \textrm{ (simplify like terms)} \\\ \therefore && 15a - 5 + 5 &= -5 + 5 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 15a &= 0 && \textrm{ (simplify)} \\\ \therefore && \frac{15a}{15} &= \frac{0}{15} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && a &= 0 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

8. Solve \(5(d - 2) = 10\) to find the value of \(d\)

\(\begin{aligned} && 5(d - 2) &= 10 \\\ \therefore && 5d - 10 &= 10 && \textrm{ (expand brackets)} \\\ \therefore && 5d - 10 + 10 &= 10 + 10 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 5d &= 20 && \textrm{ (simplify)} \\\ \therefore && \frac{5d}{5} &= \frac{20}{5} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && d &= 4 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

9. Solve \(3(d - 4) = 12\) to find the value of \(d\)

\(\begin{aligned} && 3(d - 4) &= 12 \\\ \therefore && 3d - 12 &= 12 && \textrm{ (expand brackets)} \\\ \therefore && 3d - 12 + 12 &= 12 + 12 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 3d &= 24 && \textrm{ (simplify)} \\\ \therefore && \frac{3d}{3} &= \frac{24}{3} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && d &= 8 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

10. Solve \(\frac{2a - 1}{6} = 4\) to find the value of \(a\)

\(\begin{aligned} && \frac{2a - 1}{6} &= 4 \\\ \therefore && 2a - 1 &= 24 && \textrm{ (simplify)} \\\ \therefore && 2a - 1 + 1 &= 24 + 1 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 2a &= 25 && \textrm{ (simplify)} \\\ \therefore && \frac{2a}{2} &= \frac{25}{2} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && a &= 12.5 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

11. Solve \(\frac{-2t + 1}{3} + 2 = 6\) to find the value of \(t\)

\(\begin{aligned} && \frac{-2t + 1}{3} + 2 &= 6 \\\ \therefore && \frac{-2t + 1}{3} + 2 - 2 &= 6 - 2 && \textrm{ (undo addition by subtracting)} \\\ \therefore && \frac{-2t + 1}{3} &= 4 && \textrm{ (simplify)} \\\ \therefore && 3(\frac{-2t + 1}{3}) &= 3(4) && \textrm{ (undo division by multiplying)} \\\ \therefore && -2t + 1 &= 12 && \textrm{ (simplify)} \\\ \therefore && -2t + 1 - 1 &= 12 - 1 && \textrm{ (undo addition by subtracting)} \\\ \therefore && -2t &= 11 && \textrm{ (simplify)} \\\ \therefore && \frac{-2t}{-2} &= \frac{11}{-2} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && t &= -5.5 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)

Slope-intercept form

When you graph a linear equation with two variables, it forms a straight line. The most common way to write this type of equation is in a form that clearly shows two important features of the line. This is called slope-intercept form. It is written as follows:

\[y = mx + c\]

where:

For example, the following linear equations are all in slope-intercept form: \(y = 2x + 3\)
\(y = 4x - 5\)
\(y = -5x + 1\)

The graphs of these equations are shown below. Note that each graph has a different gradient and a different y-intercept.

Graphs of the three linear equations. The graph of 2x + 3 has a gradient of 2 and a y-intercept of 3, the graph of 4x - 5 has a gradient of 4 and a y-intercept of -5, and the graph of -5x + 1 has a gradient of -5 and a y-intercept of 1.

If a linear equation in two variables is not written in slope-intercept form, you can rearrange it into this form by using the same techniques you use to solve linear equations.

Example: Rearrange the linear equation \(6x + 2y = 10\) to be in slope-intercept form.

Solution: Rearranging using the steps for solving linear equations gives:

\[\begin{aligned} && 6x + 2y &= 10 \\\ \therefore && 6x + 2y - 6x &= 10 - 6x && \textrm{ (remove } 6x \textrm{ from LHS of equation by subtracting)} \\\ \therefore && 2y &= 10 - 6x && \textrm{ (simplify)} \\\ \therefore && 2y/2 &= 10/2 - 6x/2 && \textrm{ (undo multiplication by dividing)} \\\ \therefore && y &= 5 - 3x && \textrm{ (simplify)} \\\ \therefore && y &= - 3x + 5&& \textrm{ (rearrange)} \end{aligned}\]

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Once you have worked through the example above, try some or all of the following problems that involve linear equations in slope-intercept form:

1. State the gradient and y-intercept of the straight line \(y = 3x - 4\)

The gradient of this line is \(3\) and the y-intercept is \(-4\)

2. State the gradient and y-intercept of the straight line \(y = -0.4x + 7\)

The gradient of this line is \(-0.4\) and the y-intercept is \(7\)

3. Rearrange the linear equation \(-12x = 6 - 3y\)

\(\begin{aligned} && -12x &= 6 - 3y \\\ \therefore && -12x + 3y &= 6 - 3y + 3y && \textrm{ (remove }-3y \textrm{ from RHS of equation by adding} \\\ \therefore && -12x + 3y &= 6 && \textrm{ (simplify)} \\\ \therefore && -12x + 3y + 12x &= 6 + 12x && \textrm{ (remove }-12x \textrm{ from LHS of equation by adding} \\\ \therefore && 3y &= 6 + 12x && \textrm{ (simplify)} \\\ \therefore && 3y/3 &= 6/3 + 12x/3 && \textrm{ (undo multiplication by dividing)} \\\ \therefore && y &= 4x + 2 && \textrm{ (simplify and rearrange)} \\\ \end{aligned}\)

4. Rearrange the linear equation \(4y + 20x + 8 = 0\)

\(\begin{aligned} && 4y + 20x + 8 &= 0 \\\ \therefore && 4y + 20x + 8 - 8 &= 0 - 8 && \textrm{ (remove } 8 \textrm{ from LHS of equation by subtracting} \\\ \therefore && 4y + 20x &= -8 && \textrm{ (simplify)} \\\ \therefore && 4y + 20x - 20x &= -8 - 20x && \textrm{ (remove } 20x \textrm{ from LHS of equation by subtracting} \\\ \therefore && 4y &= -8 - 20x && \textrm{ (simplify)} \\\ \therefore && 4y/4 &= -8/4 - 20x/4 && \textrm{ (undo multiplication by dividing)} \\\ \therefore && y &= -5x - 2 && \textrm{ (simplify and rearrange)} \\\ \end{aligned}\)

Quadratic equations

This page introduces the concept of quadratic equations and explains some techniques used to solve them.

In brief, it covers the following:

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What is a quadratic equation?

A quadratic equation, or second degree equation, is an algebraic equation of the form:

\[ax^2 + bx + c = 0,\]

where \(x\) is a variable and \(a\), \(b\) and \(c\) are known numbers, where \(a \neq 0\). These are referred to as coefficients of the equation. NOte that if \(a = 0\) then the equation is linear, not a quadratic.

The most basic quadratic equation occurs when \(a = 1\), \(b = 0\) and \(c = 0\). In this case the equation becomes:

\[x^2 = 0\]

From this, you can see that the value of the variable \(x\) must be \(0\), because only \(0^2 = 0\). So, the solution to this quadratic equation is \(0\).

This quadratic equation has only one solution. However, many quadratic equations have two solutions, and some have no real solutions at all. To understand why, it can be helpful to look at the graphs of quadratic equations. These graphs, shown below, are called parabolas. The points where a parabola crosses \(x\) axis (the horizontal axis) are the solutions to the equation, because these are the points where the equation equals \(0\).

The first image below shows a parabola that crosses the \(x\) axis once, so there is only one solution. The second image shows a parabola that crosses the \(x\) axis twice, so there are two solutions. The third image shows a parabola that does not cross the \(x\) axis at all, so there are no real solutions.

(Remember, the point where a graph crosses the \(x\) axis is called the \(x\) intercept.)

Graphs of three quadratic equations. The first graph intersects the x-axis once, the second graph intersects the x-axis twice and the third graph does not intersect the x-axis at all.

There are several ways to solve quadratic equations, depending on the type of equation. This page explains three different techniques.

Solving when \(b = 0\)

Probably the easiest type of quadratic equation to solve is one where \(b = 0\); that is, when the equation has the form:

\[ax^2 + c = 0\]

You can solve equations like this by following these steps:

  1. First, rearrange the equation so that \(ax^2\) is by itself on the left-hand side. Do this by adding or subtracting the constant \(c\) from both sides, just as you would for a linear equation.

  2. Next, divide both sides of the equation by \(a\) (as long as \(a \neq 0\)), again using the same method you would use for a linear equation.

  3. Then, take the square root of both sides of the equation. When you do this, remember that there are three possible outcomes:
    • there is no real solution, if you are taking the square root of a negative number, because no real nunmber squared gives a negative result;
    • there is one solution, \(x = 0\), if you are taking the square root of \(0\); or
    • there are two solutions, one positive and one negative, if you are taking the square root of a positive number. For example, \(\sqrt{100} = \pm 10\) (“plus or minus ten”), because both \(10\) and \(-10\) are square roots of \(100\).
  4. Finally, check your answer(s) by substituting each value of \(x\) back into the original equation. If the result is not correct, go back through your steps to find and fix the mistake.

Example: Solve the quadratic equation \(2x^2 - 72 = 0\)

Solution: Following the steps above, we have:

\[\begin{aligned} && 2x^2 - 72 &= 0 && \\\ \therefore && 2x^2 - 72 + 72 &= 0 + 72 && \textrm{ (undo subtraction by adding) } \\\ \therefore && 2x^2 &= 72 && \textrm{ (simplify) } \\\ \therefore && \frac{2x^2}{2} &= \frac{72}{2} && \textrm{ (divide both sides by 2) } \\\ \therefore && x^2 &= 36 && \textrm{ (simplify) } \\\ \therefore && \sqrt{x^2} &= \sqrt{36} && \textrm{ (take square root of both sides) } \\\ \therefore && x &= \pm 6 && \textrm{ (evaluate - equation is solved!) } \end{aligned}\]

Finally, substituting \(\pm 6\) back into the original equation gives \(2(\pm 6)^2 - 72 = 2(36) - 72 = 72 - 72 = 0\) as required. So the solution is correct.

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Once you have worked through the example above, have a go at using this technique to find the value of \(x\) in some or all of the following quadratic equations:

1. \(x^2 - 36 = 0\)

\[\begin{aligned} && x^2 - 36 &= 0 \\\ \therefore && x^2 - 36 + 36 &= 0 + 36 && \textrm{ (undo subtraction by adding) } \\\ \therefore && x^2 &= 36 && \textrm{ (simplify) } \\\ \therefore && \sqrt{x^2} &= \sqrt{36} && \textrm{ (take square root of both sides) } \\\ \therefore && x &= \pm6 && \textrm{ (simplify - equation is solved!) } \end{aligned}\]

2. \(-x^2 + 100 = 0\)

\[\begin{aligned} && -x^2 + 100 &= 0 \\\ \therefore && -x^2 + 100 - 100 &= 0 - 100 && \textrm{ (undo addition by subtracting) } \\\ \therefore && -x^2 &= -100 && \textrm{ (simplify) } \\\ \therefore && \frac{-x^2}{-1} &= \frac{-100}{-1} && \textrm{ (undo multiplication by dividing) } \\\ \therefore && x^2 &= 100 && \textrm{ (simplify) } \\\ \therefore && \sqrt{x^2} &= \sqrt{100} && \textrm{ (take square root of both sides) } \\\ \therefore && x &= \pm10 && \textrm{ (simplify - equation is solved!) } \end{aligned}\]

3. \(5x^2 + 30 = 0\)

\[\begin{aligned} && 5x^2 + 30 &= 0 \\\ \therefore && 5x^2 + 30 - 30 &= 0 - 30 && \textrm{ (undo addition by subtracting) } \\\ \therefore && 5x^2 &= -30 && \textrm{ (simplify) } \\\ \therefore && \frac{5x^2}{5} &= \frac{-30}{5} && \textrm{ (undo multiplication by dividing) } \\\ \therefore && x^2 &= -6 && \textrm{ (simplify) } \\\ \therefore && \sqrt{x^2} &= \sqrt{-6} && \textrm{ (take square root of both sides) } \\\ \therefore && x &\textrm{ has no real solutions } && \end{aligned}\]

Solving by factorising

Here is another way to solve a quadratic equation. This method uses factorising, where you write the equation using two sets of brackets. This method is not always easy. Sometimes it is not even possible. However, if you can see that the equation is easy to factorise, this method can help you solve it quickly. Follow these steps:

  1. Factorise the quadratic equation. You can learn how to do this in the Factorising section of this module.

  2. For the quadratic equation to equal zero, at least one of the factors must equal zero. This means you can solve the equation by setting each factor equal to zero. Then solve each one separately, the same way you would solve a linear equation.

  3. Check that your answer, or answers, are correct. To do this, put the value back into the original equation and work it out. If the answer is not correct, look back through your steps to find your mistake. Then try again.

Example: Solve the quadratic equation \(x^2 + 7x + 12 = 0\)

Solution: Following the steps above, we have:

  1. \(x^2 + 7x + 12 = 0\) can be factorised as \((x + 3)(x + 4)\)

  2. Therefore either \(x + 3 = 0\), or \(x + 4 = 0\)
    Solving \(x + 3 = 0\) gives \(x = -3\), while solving \(x + 4 = 0\) gives \(x = -4\)
    Therefore the two solutions are \(x = -3\) and \(x = -4\)

  3. Substituting these values back into the original equation gives \((-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0\) and \((-4)^2 + 7(-4) + 12 = 16 - 28 + 12 = 0\) as required.

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Once you have worked through the example above, have a go at using this technique to find the value of \(x\) in some or all of the following quadratic equations:

1. \(x^2 + 7x + 6 = 0\)

Factorising gives \((x + 6)(x + 1) = 0\)

Therefore \(x + 6 = 0\), or \(x + 1 = 0\)

Solving these equations respectively gives two solutions: \(x = -6\) or \(x = -1\)

2. \(x^2 - 4x - 5 = 0\)

Factorising gives \((x + 1)(x - 5) = 0\)

Therefore \(x + 1 = 0\), or \(x - 5 = 0\)

Solving these equations respectively gives two solutions: \(x = -1\) or \(x = 5\)

3. \(x^2 + 3x = 0\)

Factorising gives \(x(x + 3) = 0\)

Therefore \(x = 0\), or \(x + 3 = 0\)

Solving gives two solutions: \(x = 0\) or \(x = -3\)

Solving using the quadratic formula

Finally, there is one more way to solve any quadratic equation. This method uses the quadratic formula. This method will not always be the fastest way to find the answer. Because of this, it is a good idea to check first whether the other two methods will work. However, the quadratic formula works in every situation. Sometimes, it is the only method that will work.

The quadratic formula says that for a quadratic equation written as \(ax^2 + bx + c = 0\), you can find the value of \(x\) using this formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Example: Solve the quadratic equation \(2x^2 + 3x - 10 = 0\)

Solution: Following the steps above, we have:

\[\begin{aligned} x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-10)}}{2(2)} \\\ &= \frac{-3 \pm \sqrt{9 + 80}}{4} \\\ &= \frac{-3 \pm \sqrt{89}}{4} \\\ &= \frac{-3 \pm 9.43}{4} \\\ &= \frac{6.43}{4} \textrm{ or } \frac{-12.43}{4} \\\ &= 1.61 \textrm{ or } -3.11 \end{aligned}\]

Finally, substituting these values back into the original equation gives \(2(1.61)^2 + 3(1.61) - 10 \approx 0\) and \(2(-3.11)^2 + 3(-3.11) - 10 \approx 0\), as required.

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Once you have worked through the example above, have a go at solving some or all of the following quadratic equations using one of the three techniques described above:

1. \(-x^2 + 10 = 0\)

\[\begin{aligned} && -x^2 + 10 &= 0 \\\ \therefore && -x^2 + 10 - 10 &= 0 - 10 && \textrm{ (undo addition by subtracting) } \\\ \therefore && -x^2 &= -10 && \textrm{ (simplify) } \\\ \therefore && \frac{-x^2}{-1} &= \frac{-10}{-1} && \textrm{ (undo multiplication by dividing) } \\\ \therefore && x^2 &= 10 && \textrm{ (simplify) } \\\ \therefore && \sqrt{x^2} &= \sqrt{10} && \textrm{ (take square root of both sides) } \\\ \therefore && x &= \pm3.16 && \textrm{ (simplify - equation is solved!) } \end{aligned}\]

2. \(x^2 + 4x + 4 = 0\)

Factorising gives \((x + 2)^2 = 0\)

Therefore \(x + 2 = 0\)

Solving gives one solution: \(x = -2\)

3. \(4x^2 + 20 = 0\)

\[\begin{aligned} && 4x^2 + 20 &= 0 \\\ \therefore && 4x^2 + 20 - 20 &= 0 - 20 && \textrm{ (undo addition by subtracting) } \\\ \therefore && 4x^2 &= -20 && \textrm{ (simplify) } \\\ \therefore && \frac{4x^2}{4} &= \frac{-20}{4} && \textrm{ (undo multiplication by dividing) } \\\ \therefore && x^2 &= -5 && \textrm{ (simplify)} \\\ \therefore && \sqrt{x^2} &= \sqrt{-5} && \textrm{ (take square root of both sides)} \\\ \therefore && x &\textrm{ has no real solutions } && \end{aligned}\]

4. \(x^2 + 4x - 12 = 0\)

Factorising gives \((x + 6)(x - 2) = 0\)

Therefore \(x + 6 = 0\) or \(x - 2 = 0\)

Solving these equations respectively gives two solutions: \(x = -6\) or \(x = 2\)

5. \(x^2 +2x - 10 = 0\)

\[\begin{aligned} x &= \frac{-2 \pm \sqrt{2^2 - 4(1)(-10)}}{2(1)} \\\ &= \frac{-2 \pm \sqrt{4 + 40}}{2} \\\ &= \frac{-2 \pm \sqrt{44}}{2} \\\ &= \frac{-2 \pm 6.63}{2} \\\ &= \frac{4.63}{2} \textrm{ or } \frac{-8.63}{2} \\\ &= 2.32 \textrm{ or } -4.32 \end{aligned}\]

6. \(x^2 + 9x - 52 = 0\)

Factorising gives \((x + 13)(x - 4) = 0\)

Therefore \(x + 13 = 0\) or \(x - 4 = 0\)

Solving these equations respectively gives two solutions: \(x = -13\) or \(x = 4\)

Exponents & logarithms

This page explains what exponents and logarithms are. It also explains the key laws you can use to solve problems that involve them.

In brief, it covers the following:

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What are exponents?

Exponents are also called indices or powers. Theya re a short way of writing repeated multiplication. For example, \(5^4\) tells us that \(5\) needs to be multiplied by itself \(4\) times:

\[5^4 = 5 \times 5 \times 5 \times 5\]

In this example, the number \(5\) is called the base and the number \(4\) is called the exponent or power.

Here are some more examples of exponents, and how to work them out:

Example 1: Evaluate \(3^2\)

Solution: \(3^2 = 3 \times 3 = 9\)

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Example 2: Rewrite \(10 \times 10 \times 10 \times 10 \times 10 \times 2 \times 2\) using exponents.

Solution: Since \(10\) has been multiplied by itself \(5\) times and \(2\) has been multiplied by itself twice, we can rewrite this expression using exponents as \(10^5 \times 2^2\)

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Once you have worked through the examples above, have a go at some or all of these problems involving exponents:

1. Evaluate \(8^3\)

\(8^ 3 = 8 \times 8 \times 8 = 512\)

2. Evaluate \(0^3\)

\(0^3 = 0 \times 0 \times 0 = 0\)

3. Evaluate \(10^9\)

\(10^9 = 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 1 000 000 000\)

4. Rewrite \(2 \times 2 \times 2 \times 2\) using exponents.

\(2^4\)

5. Rewrite \(5 \times 5 \times 5 \times 5 \times 5\) using exponents.

\(5^5\)

6. Rewrite \(3 \times 2 \times 2 \times 2 \times 2 \times 3\) using exponents.

\(3^2 \times 2^4\) (or \(2^4 \times 3^2\))

Exponent laws

There are several ‘laws’ that make it much easier to solve problems with exponents. These laws work no matter what the base or exponent is. However, not all the laws work when the base equals zero. Because of this, we will assume the base is not zero.

To explain these laws, we will use letters (or variables) to stand in for numbers. The letters \(x\) and/or \(y\) will stand for bases that are not zero. The letters \(a\) and/or \(b\) will stand for exponents. For example:

\(x^a\) means \(x\) multiplied by itself \(a\) times

Now that we have explained what these letters mean, let’s look at the exponent laws. These laws are numbered here to make them easy to follow. However, there is no official numbering system. You may find them listed in a different order in other places.

Law 1:

This law means that a base raised to the power of 1 is simply equal to the base itself.

Law 1: \(x^1 = x\)

This law means that a base raised to the power of \(1\) is equal to the base itself.

Example: Use Law 1 to evaluate \(5^1\)

Solution: \(5^1 = 5\)

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Law 2: \(x^0 = 1\)

This law means that a base raised to the power of \(0\) is equal to \(1\).

Example: Use Law 2 to evaluate \(4^0\)

Solution: \(4^0 = 1\)

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Law 3: \(x^ax^b = x^{a+b}\)

This law means that when you multiply two exponents that have the same base you can rewrite them as one exponent. To do this, you add the powers together.

Example: Show how Law 3 works in order to evaluate \(2^32^4\)

Solution: \(2^32^4 = (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7 (= 2^{3+4})\)

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Law 4: \(\frac{x^a}{x^b} = x^{a-b}\)

This law means that when you divide two exponents that have the same base, you can rewrite them as one exponent. To do this, you subtract the powers.

Example: Show how Law 4 works in order to evaluate \(\frac{10^5}{10^3}\)

Solution: \(\frac{10^5}{10^3} = \frac{10 \times 10 \times 10 \times 10 \times 10}{10 \times 10 \times 10} = 10 \times 10 = 10^2 (=10^{5-3})\)

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Once you have worked through the examples above, have a go at using the exponent laws covered so far to evaluate some or all of the following:

1. Evaluate \(3^23^33^5\)

\(= 3^{2+3+5}\)

\(= 3^{10}\)

2. Evaluate \(\frac{5^35^4}{5^6}\)

\(= \frac{5^{3+4}}{5^6}\)

\(= \frac{5^7}{5^6}\)

\(= 5^{7-6}\)

\(= 5^1\)

\(= 5\)

3. Evaluate \(4^34^{-3}\)

\(= 4^{3+(-3)}\)

\(= 4^{3-3}\)

\(= 4^0\)

\(= 1\)

4. Evaluate \(2^32^2\)

\(= 2^{3+2}\)

\(= 2^5\)

5. Evaluate \(\frac{4^8}{4^5}\)

\(= 4^{8-5}\)

\(= 4^3\)

6. Evaluate \(\frac{5^45^2}{5^3}\)

\(= \frac{5^{4+2}}{5^3}\)

\(= \frac{5^6}{5^3}\)

\(= 5^{6-3}\)

\(= 5^3\)

Once you feel confident with the first four exponent laws, have a look at the next four:

Law 5: \(x^{-a} = \frac{1}{x^a}\)

This law means you can write a negative exponent in two different ways. You can write it as a negative exponent. Or, you can write it as \(1\) over the positive version of the exponent. This second way is called the reciprocal of the positive exponent.

Both ways are correct. However, sometimes you will be asked to use one particular way. At other times, using one way instead of the other will help you simplify an expression or solve an equation more easily.

Example 1: Use Law 5 to rewrite \(2^{-3}\) using a positive exponent.

Solution: \(2^{-3} = \frac{1}{2^3}\)

Example 2: Use Law 5 (and Law 1) to rewrite \(\frac{1}{4}\) using a negative exponent.

Solution: \(\frac{1}{4} = \frac{1}{4^1} = 4^{-1}\)

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Law 6: \((x^a)^b = x^{ab}\)

This law means that if you have a power inside brackets and another power outside the brackets, you multiply the two powes together.

Example: Show how Law 6 works in order to evaluate \((2^2)^3\)

Solution: \((2^2)^3 = (2 \times 2)^3 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 (= 2^{2 \times 3})\)

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Law 7: \((xy)^a = x^ay^a\)

This law means that when you multiply two or more exponents that have the same power, you can write them in two ways. You can write the bases together inside brackets, with one power outside the brackets. Or, you can write each exponent out separately. Sometimes you will be asked to use one particular way. At other times, using one way instead of the other will help you simplify an expression or solve an equation more easily.

Example: Show how Law 7 works in order to evaluate \((4 \times 5)^3\)

Solution: \((4 \times 5)^3 = (4 \times 5) \times (4 \times 5) \times (4 \times 5) = 4 \times 5 \times 4 \times 5 \times 4 \times 5 = 4 \times 4 \times 4 \times 5 \times 5 \times 5 = 4^35^3\)

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Law 8: \((\frac{x}{y})^a = \frac{x^a}{y^a}\)

This law is an extension of Law 7. It shows that when you divide exponents that have the same power, you can also write them in two ways. You can write the bases together inside brackets with one power outside the brackets. Or, you can write each exponent out separately.

Example: Show how Law 8 works in order to evaluate \((\frac{4}{5})^3\)

Solution: \((\frac{4}{5})^3 = \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} = \frac{4 \times 4 \times 4}{5 \times 5 \times 5} = \frac{4^3}{5^3}\)

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Once you have worked through the examples above, have a go at using the exponent laws covered so far to evaluate some or all of the following:

1. Evaluate \((5^{-2})^2\), giving your answer in terms of a positive exponent.

\(= 5^{-2 \times 2}\)

\(= 5^{-4}\)

\(= \frac{1}{5^4}\)

2. Evaluate \((3^{-4}4)^{-3}\), giving your answer in terms of positive exponents.

\(= (3^{-4})^{-3}4^{-3}\)

\(= (3^{-4 \times -3})4^{-3}\)

\(= 3^{12}4^{-3}\)

\(= \frac{3^{12}}{4^3}\)

3. Evaluate \((\frac{5}{6^{-2}})^3\), giving your answer in terms of positive exponents.

\(= \frac{5^3}{(6^{-2})^3}\)

\(= \frac{5^3}{6^{-2 \times 3}}\)

\(= \frac{5^3}{6^{-6}}\)

\(= 5^36^6\)

4. Evaluate \((10^{-3})^{-2}\)

\(= 10^{-3 \times -2}\)

\(= 10^6\)

5. Evaluate \((5^24^{-4})^5\), giving your answer in terms of positive exponents.

\(= (5^2)^5(4^{-4})^5\)

\(= 5^{2 \times 5}4^{-4 \times 5}\)

\(= 5^{10}4^{-20}\)

\(= \frac{5^{10}}{4^{20}}\)

6. Evaluate \((\frac{3^2}{4^{-3}})^{-5}\), giving your answer in terms of positive exponents.

\(= \frac{(3^2)^{-5}}{(4^{-3})^{-5}}\)

\(= \frac{3^{2 \times -5}}{4^{-3 \times -5}}\)

\(= \frac{3^{-10}}{4^{15}}\)

\(= \frac{1}{3^{10}4^{15}}\)

Once you feel confident with the first eight exponent laws, have a look at these last two:

Law 9: \(x^{\frac{1}{a}} = \sqrt[a]{x}\)

This law applies when your power is a fraction, with a numerator (the top number) \(1\) and denominator (the bottom number) \(a\). In this case you can either leave the exponent as a fraction, or you can rewrite it as the \(a^{th}\) root of the base. A root is similar to a square root, but usingnumbers other than 2, which in this case is represented by \(a\). The root tells you the number that, when multiplied by itself a certain number of times, gives you the number inside the root. For example, \(\sqrt[3]{8} = 2\), because (2 \times 2 \times 2 = 8\).

Both ways of writing this are correct. However, sometimes you will be asked to use one particular way. At other times, using one way instead of the other will help you simplify an expression or solve an equation more easily.

Example 1: Use Law 9 to rewrite \(9^{\frac{1}{2}}\) as a root of the base.

Solution: \(9^{\frac{1}{2}} = \sqrt[2]{9} = \sqrt{9}\)

Example 2: Use Law 9 to rewrite \(\textrm{ }\sqrt[3]{27}\) as an exponent with a fractional power.

Solution: \(\sqrt[3]{27} = 27^{\frac{1}{3}}\)

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Law 10: \(x^{\frac{b}{a}} = \sqrt[a]{x^b} =( \sqrt[a]{x})^b\)

This law is an extension of Law 9, and sometimes this law is given instead of Law 9. It applies when your power is a fraction, with a numerator (the top number) \(b\) and a denominator (the bottom number) \(a\). In this case, you can write the exponent as a fraction. Or, you can write it as the \(a^{th}\) root of the base, raised to the power of \(b\). This power can be placed either inside the root sign or outside it. You can try this yourself, to check that both placements give you the same answer.

All three ways of writing this are correct. However, sometimes you will be asked to use one particular way. At other times, using one way instead of the other will help you simplify an expression or solve an equation more easily.

Example 1: Use Laws 6 and 9 to show how Law 10 works to evaluate \(8^{\frac{2}{3}}\) as a root of the base, with the power inside the root sign.

Solution: \(8^{\frac{2}{3}} = 8^{2 \times \frac{1}{3}} = (8^2)^{\frac{1}{3}} = \sqrt[3]{8^2}\)

Example 2: Use Laws 6 and 9 to show how Law 10 works to evaluate \(8^{\frac{2}{3}}\) as a root of the base, with the power outside the root sign.

Solution: \(8^{\frac{2}{3}} = 8^{\frac{1}{3} \times 2} = (8^{\frac{1}{3}})^2 = (\sqrt[3]{8})^2\)

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Once you have worked through the examples above, have a go at using the exponent laws to evaluate some or all of the following:

1. Evaluate \(16^{\frac{1}{4}}\), by first converting the exponent to a root. Don’t forget you should end up with both a positive and negative answer!

\(= \sqrt[4]{16}\)

\(= \pm 2\)

2. Evaluate \(64^{\frac{1}{2}}\), by first converting the exponent to a root. Don’t forget you should end up with both a positive and negative answer!

\(= \sqrt[2]{64}\)

\(= \pm 8\)

3. Evaluate \(27^{\frac{5}{3}}\), by first converting the exponent to a root.

\(= (\sqrt[3]{27})^5\)

\(= 3^5\)

\(= 243\)

4. Evaluate \(8^{\frac{4}{3}}\), by first converting the exponent to a root.

\(= (\sqrt[3]{8})^4\)

\(= 2^4\)

\(= 16\)

What are logarithms?

Logarithms are the ‘opposite’ of exponents. This is similar to how addition and subtraction are opposites, and how multiplication and division are opposites. In other words, logarithms and exponents undo each other.

Logarithms also have a base. The logarithm of a number, using a certain base, tells you how many times you need to multiply the base by itself to get that number. For example, if you work out \(\log_b{a}\) (‘the logarithm of \(a\) with base \(b\)’) this tells you how many times you need to multiply \(b\) by itself to reach \(a\).

Keep in mind that you can only find the logarithm of positive numbers. Also, only positive bases are useful for logarithms. Because of this, we will only work with positive numbers when using logarithms.

To work out a logarithm manually, you can multiply the base by itself over and over again until you reach the correct answer.

Example: Evaluate \(\log_2{16}\)

Solution: To evaluate this you would determine how many times \(2\) needs to be multiplied by itself in order to equal \(16\). Repeated multiplication gives a solution of \(4\), since:

\[2 \times 2 \times 2 \times 2 = 2^4 = 16\]

Therefore we have \(\log_2{16} = 4\), which tells us that the logarithm of \(16\) with base \(2\) is \(4\)

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Because logarithms and exponents are opposite operations, you can rewrite one as the other. You may have already noticed this in the example above.

Specifically, if you have an equation where \(a\) is equal to the base \(b\) raised to the power \(c\) (and \(a\), \(b\) and \(c\) are all positive numbers). This equation can be rewritten so that \(c\) is equal to the logarithm of \(a\), using base \(b\). This also works the other way around, so you can rewrite the logarithm equation back into the exponent equation:

\[a = b^c \Longleftrightarrow c = \log_b{a}\]

This relationship is useful because it means logarithms can help you solve for an unknown exponent in an exponential equation. This is one of the main reasons you might need to use logarithms.

Once you have worked through the above, have a go at some or all of these problems involving logarithms:

1. Evaluate \(\log_5{125}\)

\(5 \times 5 \times 5 = 5^3 = 125\)

So \(\log_5{125} = 3\)

2. Evaluate \(\log_4{1024}\)

\(4 \times 4 \times 4 \times 4 \times 4 = 4^5 = 1024\)

So \(\log_4{1024} = 5\)

3. Evaluate \(\log_3{243}\)

\(3 \times 3 \times 3 \times 3 \times 3 = 3^5 = 243\)

So \(\log_3{243} = 5\)

4. Evaluate \(\log_2{128}\)

\(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7 = 128\)

So \(\log_2{128} = 7\)

5. Rearrange \(y = 5^x\) to give an equation for \(x\) in terms of a logarithm.

Rearranging gives \(x = \log_5{y}\)

6. Rearrange \(\log_3{x} = 5\) to give an equation for \(x\) in terms of an exponent, and hence solve for \(x\)

Rearranging gives \(x = 3^5\)

Hence \(x = 243\)

7. Rearrange \(\log_7{x} = 2\) to give an equation for \(x\) in terms of an exponent, and hence solve for \(x\)

Rearranging gives \(x = 7^2\)

Hence \(x = 49\)

Logarithms can have many different bases, as shown in the examples above. However, there are two bases that are used more often than any other. Scientific calculators have buttons for working out logarithms using these two bases. The first of these bases is \(10\), and it is used so often that if you see a logarithm written without a base shown, you can usually assume the base is \(10\). For example:

\[\log{100000} = \log_{10}{100000}\]

On a scientific calculator, you can work out a logarithm with base \(10\) by pressing the ‘\(\log\)’ button.

Example: Evaluate \(\log{100000}\)

Solution: This logarithm can be evaluated on a calculator, for example by entering \(100000\) and then pressing ‘\(\log\)’ (although some calculators may require this in a different order). The result will be \(5\), indicating that \(\log{100000} = 5\)

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The other base that is commonly used for logarithms is a special number called Euler’s number. This number is shown using the letter \(e\).

\(e\) is an irrational number, which means its decimal value goes on forever without repeating. It is a very important number in mathematics, economics and finance, and its first \(20\) digits are:

\[2.7182818284590452353…\]

A logarithm with base \(e\) is called a natural logarithm. Instead of writing this as log, it is usually written using the symbol \(\ln\). For example:

\[\ln{7.389} = \log_e{7.389}\]

On a scientific calculator, you can work out a natural logarithm by pressing the ‘\(\ln\)’ button.

Example: Evaluate \(\ln{7.389}\)

Solution: This logarithm can be evaluated on a calculator, for example by entering \(7.389\) and then pressing ‘\(\ln\)’ (although some calculators may require this in a different order). The result will be approximately \(2\), indicating that \(\ln{7.389} \approx 2\)

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Once you have worked through the above, have a go at some or all of these problems involving evaluating logarithms on a calculator:

1. Evaluate \(\log{500}\)

Evaluating this logarithm using a calculator gives \(2.70\)

2. Evaluate \(\ln{200000}\)

Evaluating this logarithm using a calculator gives \(12.21\)

3. Evaluate \(\log{10000}\)

Evaluating this logarithm using a calculator gives \(4\)

4. Evaluate \(\ln{550}\), rounding to two decimal places

Evaluating this logarithm using a calculator gives \(6.31\)

5. Rearrange \(y = e^x\) to give an equation for \(x\) in terms of a natural logarithm

Rearranging gives \(x = \log_e{y} = \ln{y}\)

6. Rearrange \(400 = e^x\) to give an equation for \(x\) in terms of a natural logarithm, and hence solve for \(x\)

Rearranging gives \(x = \log_e{400} = \ln{400}\)

Hence \(x = 5.99\)

7. Rearrange \(\log{x} = 3\) to give an equation for \(x\) in terms of an exponent, and hence solve for \(x\)

Rearranging gives \(x = 10^3\)

Hence \(x = 1000\)

Lastly, it’s important to know that you can still work out logarithms that use a base other than \(10\) or \(e\), even though your calculator doesn’t have a button for them. To do this, you first need to use the change of base rule.

This rule lets you rewrite a logarithm using a different base of your choosing. This new base is shown as \(c\) below, and it can be any positive number you like, but its usually easiest to choose \(10\) or \(e\), since these are the two bases your calculator can work with directly. The rule is:

\[\log_b{a} = \frac{\log_c{a}}{\log_c{b}}\]

Example: Evaluate \(\log_4{12}\) on a calculator by first applying the change of base rule.

Solution: Using the change of base rule to convert both logarithms to base \(10\) gives the following (note that you could also convert both logarithms to base \(e\), which would give the same result):

\[\begin{aligned} \log_4{12} &= \frac{\log_{10}{12}}{\log_{10}{4}} \\\ &= \frac{\log{12}}{\log{4}} \\\ &= \frac{1.079}{0.602} \\\ &= 1.792 \end{aligned}\]

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Once you have worked through the above, have a go at some or all of these problems involving applying the change of base rule in order to evaluate logarithms on a calculator:

1. Evaluate \(\log_6{20}\)

\[\begin{aligned} \log_6{20} &= \frac{\log_{10}{20}}{\log_{10}{6}} \\\ &= \frac{\log{20}}{\log{6}} \\\ &= \frac{1.301}{0.778} \\\ &= 1.672 \end{aligned}\]

2. Evaluate \(\log_7{114}\)

\[\begin{aligned} \log_7{114} &= \frac{\log_{10}{114}}{\log_{10}{7}} \\\ &= \frac{\log{114}}{\log{7}} \\\ &= \frac{2.057}{0.845} \\\ &= 2.434 \end{aligned}\]

3. Evaluate \(\log_5{200}\)

\[\begin{aligned} \log_5{200} &= \frac{\log_{10}{200}}{\log_{10}{5}} \\\ &= \frac{\log{200}}{\log{5}} \\\ &= \frac{2.301}{0.699} \\\ &= 3.292 \end{aligned}\]

4. Evaluate \(\log_9{1500}\)

\[\begin{aligned} \log_9{1500} &= \frac{\log_{10}{1500}}{\log_{10}{9}} \\\ &= \frac{\log{1500}}{\log{9}} \\\ &= \frac{3.176}{0.954} \\\ &= 3.328 \end{aligned}\]

5. Evaluate \(\log_3{2260}\)

\[\begin{aligned} \log_3{2260} &= \frac{\log_{10}{2260}}{\log_{10}{3}} \\\ &= \frac{\log{2260}}{\log{3}} \\\ &= \frac{3.354}{0.477} \\\ &= 7.030 \end{aligned}\]

Logarithm laws

Just like with exponents, there are some ‘laws’ that make it much easier to solve problems involving logarithms. We will use variables to write these laws below. The letters \(a\), \(b\) and \(c\) are used to represent any positive numbers. The five most commonly used logarithm laws are:

Law 1: \(\log_b{b} = 1\)

This law says that if the number inside the logarithm is the same as the base, the answer is always \(1\).

Example: Use Law 1 to evaluate \(\log_{10}{10}\)

Solution: \(\log_{10}{10} = 1\)

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Law 2: \(\log_b{1} = 0\)

Law 2 says that the logarithm of 1 is equal to zero, no matter waht the base is.

Example: Use Law 2 to evaluate \(\log_{10}{1}\)

Solution: \(\log_{10}{1} = 0\)

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Law 3: \(\log_b{ac} = \log_b{a} + \log_b{c}\)

This law says that if you are adding two logarithms that have the same base, you can combine them into one logarithm. To do this, multiply the two numbers together and keep the same base.

Example: Show how Law 3 works in order to evaluate \(\log_2{(4 \times 8)}\)

Solution: \(\log_2{(4 \times 8)} = \log_2{4} + \log_2{8}\textrm{ }[= 2 + 3 = 5 = \log_2{32}]\)

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Law 4: \(\log_b{\frac{a}{c}} = \log_b{a} - \log_b{c}\)

This law says that if you are subtracting one logarithm from another and they both have the same base, you can combine them into one logarithm. To do this, divide the first number by the second and keep the same base.

Example: Show how Law 4 works in order to evaluate \(\log_2{\frac{64}{4}}\)

Solution: \(\log_2{\frac{64}{4}} = \log_2{64} - \log_2{4}\textrm{ }[= 6 - 2 = 4 = \log_2{16}]\)

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Law 5: \(\log_b{a^c} = c \log_b{a}\)

This law says that if you have the logarithm of a number that is raised to a power, this is the same as multiplying the logarithm by that power.

Example: Show how Law 5 works in order to evaluate \(\log_2{8^2}\)

Solution: \(\log_2{8^2} = 2 \log_2{8}\textrm{ }[= 2(3) = 6 = \log_2{64}]\)

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Once you have worked through the examples above, have a go at using the logarithm laws to evaluate some or all of the following:

1. Evaluate \(\log_{15}{15} + \log_3{3}\)

\(\log_{15}{15} + \log_3{3} = 1 + 1 = 2\)

2. Evaluate \(\log_{100}{100} + log_{15}{1}\)

\(\log_{100}{100} + log_{15}{1} = 1 + 0 = 1\)

3. Evaluate \(\log{(\frac{1}{100})}\)

\(\log{(\frac{1}{100})} = \log{1} - \log{100} = 0 - 2 = -2\)

4. Evaluate \(\log_6{4} + \log_6{54} + \log_7{1}\)

\(\log_{6}4 + \log_6{54} + 0 = \log_{6}4 + \log_6{54} = \log_{6}{(4 \times 54)} = \log_6{216} = 3\)

5. Evaluate \(2\log_8{2} + \log_8{128}\)

\(2\log_8{2} + \log_8{128} = \log_8{(2^2)} + \log_8{128} = \log_8{4} + \log_8{128} = \log_8{(4 \times 128)} = \log_8{512} = 3\)

6. Evaluate \(\log{10^5} - \log{10^2}\)

\(\log{(10^5)} - \log{(10^2)} = 5\log{10} - 2\log{10} = 5 - 2 = 3\)

7. Evaluate \(\log_2{40} + 3\log_2{6} - \log_2{270}\)

\(\log_2{40} + 3\log_2{6} - \log_2{270} = \log_2{40} + \log_2{(6^3)} - \log_2{270} = \log_2{40} + \log_2{216} - \log_2{270} = \log_2{(\frac{40 \times 216}{270})} = \log_2{32} = 5\)

Functions

This page introduces the idea of a function and looks at some common types, including linear, quadratic, exponential and logarithmic functions.

In brief, it covers the following:

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What is a function?

A function is a relationship between input values and output values. In this relationship, every input value has exactly one output value that matches it. You can picture this relationship like this:

Two ovals, one labelled 'input' and one labelled 'output'. Both ovals contain small black circles representing values. The four values in the 'input' oval map to values in the 'output' oval, with two of them mapping to the same output value.

In this image, the group of input values is called the domain of the function. The group of all the output values that are possible is called the codomain. The group of output values that the input values actually match up to is called the range.

It’s fine for more than one input value to match up to the same output value, as shown in the image above. However, one input value can never match up to more than one output value. If this happens, the relationship is not a function.

As an example of a function, have a look at the linear relationship:

\[y = 2x + 3\]

This function takes a number \(x\) and changes it by multiplying it by \(2\) and then adding \(3,\). This gives you an output value called \(y\). Here are some examples of these input and output values:

Input value (x) Output value (y)
-1 1
0 3
1 5
2 7

The function \(y = 2x + 3\) works for all real numbers. However, this is not always the case for every relationship. Sometimes in order for a relationship to count as a function, you need to limit domain and/or the codomain.

For example, lok at the relationship \(y = \sqrt{x}\). This relationship does not give any real output values for negative input values. It also gives two output values for each positive input value. For example, the square root of \(9\) could be either \(-3\) or \(3\). Becuase of this, the relationship is only a function if both the domain and codomain are limited to non-negative real numbers (zero or higher).

If you have a graph of a relationship, you can check whether it is a function by using the vertical line test. This test says that the relationship is a only a function if every vertical line crosses the graph no more than once.

For example, look at the relationship on the left: \(y = 2x + 3\). This is a function, because there is no wat to draw a vertical line that crosses the graph more than once. However, the relationship on the right is not a function. This is because there is at least one point (in fact, there are many points) where a vertical line crosses the graph twice.

Plots of two relationships. The first relationship is a function as a vertical line at any point only intersects the graph once. The second relationship is not a function as a vertical line intersects the graph twice at some points.

Check your understanding of functions by completing the following activity:

H5P interactive activity

Function names

A function can simply be written using \(x\) and \(y\) (or other variable names depending on the situation), like the function \(y = 2x + 3\) that we saw earlier. However, functions are usually given a name. The most common name used is \(f\).

For example, the linear function \(y = 2x + 3\) could be written as:

\[f(x) = 2x + 3\]

In this example:

For example, if we use an input value of \(x = 2\) for this function, we can write the following (note that this matches the \(y\) value shown in the table above):

\[f(2) = 7\]

Check your understanding of function names by completing the following activity:

1. A function takes an input value \(z,\) squares it and then adds \(7.\) Write this as a function named \(f\)

\(f(z) = z^2 + 7\)

2. If \(f(x) = 3x - 5,\) find \(f(10)\)

\(f(10) = 25\)

Types of functions

There are many different types of functions, Here, we will look at four types that have been covered in this module. These are:

Check your understanding of different types of functions by completing the following activity:

1. \(g(x) = 10^x\) is what type of function?

This is an exponential function.

2. \(f(z) = z^2 + 3\) is what type of function?

This is a quadratic function.

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