This note covers the laws of exponents, including negative and fractional powers, and the idea of a logarithm as the inverse of an exponent. It shows how to solve equations where the unknown is in the power, which is needed for growth, decay and compound-interest models.
Key ideas
What is SL and what is HL. Integer exponents, the idea of a logarithm, base 10 and base \(e\) logarithms, and solving equations with the GDC are SL and HL. Fractional exponents, the laws of logarithms, change of base and solving exponential equations by algebra with logarithms are HL only. Where this note uses them, they are marked HL.
Laws of exponents. For \(a\gt0\) and any integer \(m, n\) (any rational \(m, n\) at HL):
- \(a^{m}\times a^{n}=a^{m+n}\)
- \(\dfrac{a^{m}}{a^{n}}=a^{m-n}\)
- \((a^{m})^{n}=a^{mn}\)
- \((ab)^{n}=a^{n}b^{n}\)
- \(a^{0}=1\) and \(a^{-n}=\dfrac{1}{a^{n}}\)
- (HL) \(a^{1/n}=\sqrt[n]{a}\) and \(a^{m/n}=\left(\sqrt[n]{a}\right)^{m}\)
(HL) For example \(16^{3/4} = (\sqrt[4]{16})^{3} = 2^{3} = 8\), and \(27^{-2/3} = \dfrac{1}{(\sqrt[3]{27})^{2}} = \dfrac{1}{9}\). Taking the root first keeps the numbers small.
Logarithms. A logarithm answers the question "what power gives this number?" The statement
\[ a^{x}=b \iff x=\log_{a}b \qquad (a\gt0,\ a\ne1,\ b\gt0) \]means that \(\log_{2}8=3\) because \(2^{3}=8\). Two bases have their own keys on a calculator. \(\log_{10}x\) is written \(\lg x\) or \(\log x\), and \(\log_{e}x\) is written \(\ln x\), where \(e\approx2.718\) is the base of natural growth. Because logs undo powers, \(a^{\log_{a}x}=x\) and \(\log_{a}a^{x}=x\). You cannot take the logarithm of zero or a negative number.
Laws of logarithms (HL). For positive \(x\) and \(y\):
- \(\log_{a}(xy)=\log_{a}x+\log_{a}y\)
- \(\log_{a}\!\left(\dfrac{x}{y}\right)=\log_{a}x-\log_{a}y\)
- \(\log_{a}(x^{k})=k\log_{a}x\)
The third law is the one that brings an unknown power down to ground level.
Solving \(a^{x}=b\). At SL use the equation solver or graph both sides on your GDC. At HL you can also take logs of both sides, \(x\log a=\log b\) so \(x=\dfrac{\log b}{\log a}\) (this is the change-of-base idea and works with \(\lg\) or \(\ln\)), or write \(x=\log_{a}b\) directly. You can also use the equation solver or graph both sides on your GDC and find the intersection. For equations of the form \(c\times a^{x}=b\), divide by \(c\) first so that the power is alone.
Equations with logs (HL). Combine logs into one using the laws, convert into a power equation, solve, and then check every solution. A solution is rejected if it puts a zero or negative number inside a logarithm.
In context, growth and decay are modelled by \(y=y_0\,a^{t}\) or \(y=y_0e^{kt}\). Logarithms allow you to find the time \(t\) at which a target is reached.
Why logarithms are useful. Powers grow very quickly, so quantities such as sound intensity, earthquake energy and acidity span huge ranges. Logarithms compress these ranges into small numbers. They are also the tool that turns a multiplication into an addition. If a value is multiplied by 10 each time, its \(\log_{10}\) goes up by exactly 1.
Useful values to recognise. Knowing some powers helps you check answers quickly: \(\log_{10}1000=3\), \(\log_{2}64=6\), \(\log_{3}\dfrac19=-2\), \(\ln e=1\) and \(\log_{a}1=0\) for any base. A logarithm is negative when the number inside is between 0 and 1.
Working with exponents in algebra. When simplifying, give every answer with positive powers unless told otherwise, and write roots as fractional powers before you apply the laws. For example \(\dfrac{(2x^{3})^{2}}{x^{5}}=\dfrac{4x^{6}}{x^{5}}=4x\). Apply the power to the number as well as to the letter, and do not forget that \((2x^{3})^{2}\) has \(2^{2}=4\), not \(2\).
Checking a log solution. Substitute your answer back into the original equation on your GDC. If the calculator returns an error, or a value that does not match the other side, the solution is not valid.
Choosing a method for \(a^{x}=b\). If \(b\) is an exact power of \(a\) (for example \(3^{x}=81\)), write both sides with the same base and compare exponents. Otherwise use logarithms or the graphical method on your GDC.
The number \(e\) and natural logarithms. The function \(y=e^{x}\) is the exponential whose growth rate at any point equals its own value. Models such as \(N=N_0e^{kt}\) use it for continuous growth (\(k\gt0\)) or decay (\(k\lt0\)). The inverse of \(e^{x}\) is \(\ln x\), so \(\ln e^{x}=x\) and \(e^{\ln x}=x\). To find \(t\) in \(N_0e^{kt}=N\), divide by \(N_0\), take \(\ln\) of both sides and divide by \(k\). On your GDC the \(e^{x}\) function is usually the second function of the \(\ln\) key.
Half-life style questions (HL method). For decay with ratio \(a\) per step, the time for a quantity to halve solves \(a^{t}=\tfrac12\), so \(t=\dfrac{\ln(1/2)}{\ln a}\). A value of \(a\) between 0 and 1 gives a positive \(t\), because both logarithms are negative.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(a^{m}a^{n}=a^{m+n}\), \(\ \dfrac{a^{m}}{a^{n}}=a^{m-n}\) | Combining powers of the same base. |
| \((a^{m})^{n}=a^{mn}\) | A power of a power. |
| \(a^{-n}=\dfrac{1}{a^{n}}\), \(\ a^{1/n}=\sqrt[n]{a}\) | Negative and fractional exponents. |
| \(a^{x}=b\iff x=\log_{a}b\) | Definition of a logarithm. |
| \(\log_{a}xy=\log_{a}x+\log_{a}y\) | Product law. |
| \(\log_{a}\dfrac{x}{y}=\log_{a}x-\log_{a}y\) | Quotient law. |
| \(\log_{a}x^{k}=k\log_{a}x\) | Power law. Used to solve for an unknown exponent. |
| \(\log_{a}b=\dfrac{\ln b}{\ln a}\) | Change of base. Lets you use a calculator for any base. |
Worked examples
Example 1 (HL, because of the fractional powers) Evaluate \(81^{-1/4}\times 4^{3/2}\).
\(81^{-1/4}=\dfrac{1}{81^{1/4}}=\dfrac{1}{\sqrt[4]{81}}=\dfrac{1}{3}\).
\(4^{3/2}=(\sqrt{4})^{3}=2^{3}=8\).
Multiply: \(\dfrac{1}{3}\times8=\dfrac{8}{3}\).
Answer: \(\dfrac{8}{3}\) (\(=2.67\) to 3 s.f.).
Example 2 A town of 1200 people grows by 3.5% per year, so the population after \(t\) years is \(P=1200\times1.035^{t}\). Find when the population first reaches 2000. (At SL, use the GDC intersection; the log method below is the HL route.)
Set \(1200\times1.035^{t}=2000\).
Divide by 1200: \(1.035^{t}=\dfrac{2000}{1200}=\dfrac{5}{3}\).
Take natural logs of both sides: \(t\ln1.035=\ln\dfrac{5}{3}\).
\(t=\dfrac{\ln(5/3)}{\ln1.035}=\dfrac{0.51083}{0.034401}=14.849\ldots\)
Because \(t\) is in whole years of growth steps, the population exceeds 2000 during the 15th year.
Answer: \(t = 14.8\) years (to 3 s.f.), so the population passes 2000 during the 15th year.
Example 3 (HL, because it uses the laws of logarithms) Solve \(\ln x+\ln(x-2)=\ln15\).
Both logs must have positive inputs, so \(x\gt2\).
Combine with the product law: \(\ln\bigl(x(x-2)\bigr)=\ln15\).
So \(x(x-2)=15\), giving \(x^{2}-2x-15=0\), which factorises as \((x-5)(x+3)=0\).
\(x=5\) or \(x=-3\). Since \(x\gt2\), reject \(x=-3\) because \(\ln(-3)\) does not exist.
Answer: \(x=5\).
Common errors
- Common error: writing \(\log(x+y)=\log x+\log y\). The product law needs \(\log(xy)\).
- Common error: treating \(a^{-n}\) as a negative number. \(2^{-3}=\dfrac18\), not \(-8\).
- Common error: writing \(\dfrac{\log b}{\log a}=\log b-\log a\). A quotient of logs is not a log of a quotient.
- Common error: forgetting to divide by the coefficient first when solving \(c\times a^{x}=b\).
- Common error: keeping a solution that puts a negative number inside a log.
Exam tip: when the unknown is in the power, isolate the power on one side first, then take a logarithm. Show the line \(x=\dfrac{\log b}{\log a}\) before using your GDC.
Check yourself
- Solve \(2e^{0.4t}=18\).
Answer
\(e^{0.4t}=9\), so \(0.4t=\ln9\) and \(t=\dfrac{\ln 9}{0.4}=5.49\) (3 s.f.). - Find \(x\) if \(\log_{10}x=2.5\).
Answer
\(x=10^{2.5}=316\) (3 s.f.). - Solve \(7\times2^{x}=100\).
Answer
\(2^{x}=\dfrac{100}{7}\), so \(x=\dfrac{\ln(100/7)}{\ln2}=3.84\) (3 s.f.). - (HL) Simplify \(\log_{2}40-\log_{2}5\).
Answer
\(\log_{2}\dfrac{40}{5}=\log_{2}8=3\).