This page details the basics of working with exponents and logarithms. In particular, the page covers the following (use the drop-down menu above to jump to a different section as required):

  • What are exponents?
  • Exponent laws
  • What are logarithms?
  • Logarithm laws

What are exponents?

Exponents, otherwise known as indices or powers, are used as a shorthand notation for repeated multiplication. For example, \(5^4\) indicates that \(5\) needs to be multiplied by itself \(4\) times:

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

In this case we say that the number \(5\) is the base and the number \(4\) is the exponent or power. Some other examples of exponents, and how they are evaluated, are as follows:

Example 1: Evaluate \(3^2\)

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

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

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

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

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

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

\(2^4\)

\(5^5\)

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

Exponent laws

There are various ‘laws’ that will make evaluating problems involving exponents much easier. Since these laws apply regardless of the value of the base and exponent, we will use variables when detailing them (the exception is that not all the laws apply when the base equals \(0\), so we will assume a non-zero base). In particular, the variables \(x\) and/or \(y\) will be used to represent non-zero bases, and the variables \(a\) and/or \(b\) will be used to represent exponents. For example:

\(x^a\) represents \(x\) being multiplied by itself \(a\) times

Now that we have defined our variables, let’s look at the exponents laws (while these are numbered here, note that there is no ‘set’ numbering system, and you may find them ordered differently in different places):

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

In other words, a base raised to the power of \(1\) is just equal to the value of the base.

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

Solution: \(5^1 = 5\)

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

In other words, a base raised to the power of \(0\) is just equal to \(1\).

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

Solution: \(4^0 = 1\)

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

In other words, when you multiply two exponents with the same base you can rewrite it as a single exponent with the powers added together.

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

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

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

In other words, when you divide two exponents with the same base you can rewrite it as a single exponent with the powers subtracted.

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

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

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:

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

\(= 3^{10}\)

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

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

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

\(= 5^1\)

\(= 5\)

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

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

\(= 4^0\)

\(= 1\)

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

\(= 2^5\)

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

\(= 4^3\)

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

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

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

\(= 5^3\)

Once you feel you have a handle on the first four exponent laws, have a look at these next four:

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

In other words, you can write negative exponents in two different ways: either using a negative exponent; or as \(1\) over the positive version of the exponent (i.e. as the reciprocal of the positive exponent). Both are correct, but sometimes one might be specified over the other, or sometimes writing it in a particular way will better help you simplify an expression or solve an equation.

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

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

In other words, any powers inside of brackets just need to be multiplied by any powers outside of brackets.

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

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

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

In other words, when you multiply two (or more) exponents with the same power together, you can either write the bases inside brackets raised to a single power outside of the brackets, or you can write each of the exponents out separately. Again sometimes one way might be specified over the other, or sometimes writing it in a particular way will better help you simplify an expression or solve an equation.

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

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

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

This is really just an extension of the previous law, and shows that if you have exponents with the same power divided by each other, again you may wish to have them raised to a single power outside of brackets, or to write each of the exponents out separately.

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

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

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:

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

\(= 5^{-4}\)

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

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

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

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

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

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

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

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

\(= 5^36^6\)

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

\(= 10^6\)

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

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

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

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

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

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

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

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

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

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

In other words, when your power is a fraction with numerator (top number) \(1\) and denominator (bottom number) \(a\), you can either just have a fractional exponent, or you can instead write it as the \(a^{th}\) root of the base (the \(a^{th}\) root is just an extension of the square root, and is the number that when multiplied by itself \(a\) times will give the number inside the root; e.g. \(\sqrt[3]{8} = 2\), since \(2 \times 2 \times 2 = 8\)). Either is correct, but sometimes one way might be specified over the other, or sometimes writing it in a particular way will better help you simplify an expression or solve an equation.

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

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

This law is an extension of the previous law (and sometimes only this law might be specified), and says that when your power is a fraction with numerator (top number) \(b\) and denominator (bottom number) \(a\), you can either just have a fractional exponent, or you can instead write it as the \(a^{th}\) root of the base to the power of \(b\). Further, this power can be written in one of two ways; either inside the root sign or outside (try it yourself and check that they given the same value!). Any one of the three ways is correct, but sometimes one way might be specified over the other, or sometimes writing it in a particular way will better help you simplify an expression or solve an equation.

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

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

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

\(= \pm 2\)

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

\(= \pm 8\)

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

\(= 3^5\)

\(= 243\)

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

\(= 2^4\)

\(= 16\)

What are logarithms?

Logarithms are the ‘opposite’ of exponents, in the same way as addition and subtraction are opposites and multiplication and division are opposites; that is, the two operations undo each other.

Logarithms also have bases, and the logarithm of a number with a certain base tells you how many times you need to multiply the base by itself in order to obtain that number. For example, evaluating \(\log_b{a}\) (‘the logarithm of \(a\) with base \(b\)’) tells you how many times you need to multiply \(b\) by itself in order to obtain \(a\) (note however that you can only have logarithms of positive numbers, and that only positive bases are typically useful, so we will only consider positive numbers when dealing with logarithms).

In order to evaluate a logarithm manually, you can perform repeated multiplication of the base until the correct answer is obtained.

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 \(4\)

Since logarithms and exponents are opposite operations, it follows that one can be rearranged as the other, and you may have noticed this in the example above. In particular, if you have an equation that sets \(a\) equal to an exponent with base \(b\) and power \(c\) (where all three are positive numbers), it can be rearranged to an equation that sets \(c\) equal to the logarithm of \(a\) with base \(b\) (and vice versa):

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

This relationship means that logarithms can be used to solve for an unknown exponent in an exponential equation - which is one of the reasons you may need to use them.

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

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

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

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

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

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

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

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

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

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

Rearranging gives \(x = 3^5\)

Hence \(x = 243\)

Rearranging gives \(x = 7^2\)

Hence \(x = 49\)

While logarithms can have a range of different bases, as in the examples above, there are two bases that are used most commonly with logarithms - and there are buttons on a scientific calculator for evaluating these. The first of these bases is \(10\), and in fact if you see a logarithm written without any base at all you can usually assume the base to be \(10\). For example:

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

The button for evaluating a log with base \(10\) on a scientific calculator is 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 \(\log{100000} = 5\)

The other commonly used base for logarithms is a special number known as Euler’s number, which is denoted by the letter \(e\). \(e\) is an irrational number that is very important in mathematics, economics and finance, and its first \(20\) digits are:

\[2.7182818284590452353…\]

A logarithm with base \(e\) is referred to as a natural logarithm, and is generally denoted by \(\ln\). For example:

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

The button for evaluating a log with base \(e\) on a scientific calculator is 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 \(\ln{7.389} \approx 2\)

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

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

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

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

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

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

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

Hence \(x = 5.99\)

Rearranging gives \(x = 10^3\)

Hence \(x = 1000\)

Lastly, note that you can also evaluate logarithms with bases other than \(10\) or \(e\) on a scientific calculator by first applying the change of base rule. This rule is as follows (where \(c\) is a positive number of your choosing; typically either \(10\) or \(e\)):

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

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{eqnarray} \log_4{12} &=& \frac{\log_{10}{12}}{\log_{10}{4}}\ &=& \frac{\log{12}}{\log{4}}\ &=& \frac{1.079}{0.602}\ &=& 1.792 \end{eqnarray}

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:

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

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

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

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

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

Logarithm laws

Just like with exponents, there are some ‘laws’ that will make evaluating problems involving logarithms much easier. Again we will use variables when detailing these, and note in particular that \(a\), \(b\) and \(c\) will be used to represent any positive numbers. The five most commonly used logarithm laws are as follows:

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

In other words, a logarithm of a number where the base equals that number is just equal to \(1\).

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

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

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

In other words, the logarithm of 1 is equal to zero, regardless of the base.

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

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

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

In other words, when two (or more) logarithms with the same base are added together, you can instead write it as a single logarithm with the same base and the values multiplied.

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

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

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

In other words, when one logarithm is subtracted from another and they have the same base, you can instead write it as a single logarithm with the same base and the first value divided by the second.

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

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

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

In other words, when you have the logarithm of some number to a power, this is the same as the logarithm multiplied by that power.

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

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

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

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

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

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

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

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

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

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