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:
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: \(3^2 = 3 \times 3 = 9\)
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 \(2^4 \times 3^2\))
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: \(5^1 = 5\)
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: \(4^0 = 1\)
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: \(2^32^4 = (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7 (= 2^{3+4})\)
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: \(\frac{10^5}{10^3} = \frac{10 \times 10 \times 10 \times 10 \times 10}{10 \times 10 \times 10} = 10 \times 10 = 10^2 (=10^{5-3})\)
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 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: \(\frac{1}{4} = \frac{1}{4^1} = 4^{-1}\)
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: \((2^2)^3 = (2 \times 2)^3 = (2 \times 2) \times (2 \times 2) \times (2 \times 2) = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 (= 2^{2 \times 3})\)
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: \((4 \times 5)^3 = (4 \times 5) \times (4 \times 5) \times (4 \times 5) = 4 \times 5 \times 4 \times 5 \times 4 \times 5 = 4 \times 4 \times 4 \times 5 \times 5 \times 5 = 4^35^3\)
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: \((\frac{4}{5})^3 = \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} = \frac{4 \times 4 \times 4}{5 \times 5 \times 5} = \frac{4^3}{5^3}\)
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 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: \(\sqrt[3]{27} = 27^{\frac{1}{3}}\)
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: \(8^{\frac{2}{3}} = 8^{\frac{1}{3} \times 2} = (8^{\frac{1}{3}})^2 = (\sqrt[3]{8})^2\)
Once you have worked through the examples above, have a go at using the exponent laws to evaluate some or all of the following:
\(= \sqrt[4]{16}\)
\(= \pm 2\)
\(= \sqrt[2]{64}\)
\(= \pm 8\)
\(= (\sqrt[3]{27})^5\)
\(= 3^5\)
\(= 243\)
\(= (\sqrt[3]{8})^4\)
\(= 2^4\)
\(= 16\)
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 \(4\)
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:
\(5 \times 5 \times 5 = 5^3 = 125\)
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 \(x = 49\)
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 \(\log{100000} = 5\)
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 \(\ln{7.389} \approx 2\)
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, 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):
\[\begin{aligned} \log_4{12} &= \frac{\log_{10}{12}}{\log_{10}{4}} \\\ &= \frac{\log{12}}{\log{4}} \\\ &= \frac{1.079}{0.602} \\\ &= 1.792 \end{aligned}\]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:
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: \(\log_{10}{10} = 1\)
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: \(\log_{10}{1} = 0\)
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: \(\log_2{(4 \times 8)} = \log_2{4} + \log_2{8}\textrm{ }[= 2 + 3 = 5 = \log_2{32}]\)
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: \(\log_2{\frac{64}{4}} = \log_2{64} - \log_2{4}\textrm{ }[= 6 - 2 = 4 = \log_2{16}]\)
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: \(\log_2{8^2} = 2 \log_2{8}\textrm{ }[= 2(3) = 6 = \log_2{64}]\)
Once you have worked through the examples above, have a go at using the logarithm laws to evaluate some or all of the following:
\(\log_{15}{15} + \log_3{3} = 1 + 1 = 2\)
\(\log_{100}{100} + log_{15}{1} = 1 + 0 = 1\)
\(\log{(\frac{1}{100})} = \log{1} - \log{100} = 0 - 2 = -2\)
\(\log_{6}4 + \log_6{54} + 0 = \log_{6}4 + \log_6{54} = \log_{6}{(4 \times 54)} = \log_6{216} = 3\)
\(2\log_8{2} + \log_8{128} = \log_8{(2^2)} + \log_8{128} = \log_8{4} + \log_8{128} = \log_8{(4 \times 128)} = \log_8{512} = 3\)
\(\log{(10^5)} - \log{(10^2)} = 5\log{10} - 2\log{10} = 5 - 2 = 3\)
\(\log_2{40} + 3\log_2{6} - \log_2{270} = \log_2{40} + \log_2{(6^3)} - \log_2{270} = \log_2{40} + \log_2{216} - \log_2{270} = \log_2{(\frac{40 \times 216}{270})} = \log_2{32} = 5\)