This note covers exponential functions, logarithms as their inverses, the laws of logarithms, and how to solve equations where the unknown is in a power. These ideas model growth and decay in money, populations and radioactivity, so they appear in many other topics.
Key ideas
An exponential function has the form \(f(x)=a^x\) with \(a\gt 0\) and \(a\ne 1\). The variable is in the power. Its graph always passes through \((0,1)\) and has the \(x\)-axis (\(y=0\)) as a horizontal asymptote. If \(a\gt 1\) the graph rises, which is growth. If \(0\lt a\lt 1\) it falls, which is decay.
The number \(e\approx 2.71828\) is a special base. The function \(e^x\) is its own gradient, which makes it the natural choice for continuous growth and decay. Any exponential can be written in base \(e\) using \(a^x=e^{x\ln a}\).
A logarithm undoes an exponential: \(\log_a x=y\) means \(a^y=x\). So \(y=\log_a x\) is the reflection of \(y=a^x\) in the line \(y=x\). This gives these facts about the graph of \(y=\log_a x\):
- It is only defined for \(x\gt 0\).
- It passes through \((1,0)\), because \(a^0=1\).
- It has the \(y\)-axis (\(x=0\)) as a vertical asymptote.
The natural logarithm is \(\ln x=\log_e x\). Because they are inverses, \(a^{\log_a x}=x\) for \(x\gt 0\) and \(\log_a(a^x)=x\) for all \(x\).
The laws of logarithms come directly from the laws of indices. To solve an equation with the unknown in an exponent, take logarithms of both sides, then use the power law to bring the exponent down. To solve an equation containing logarithms, combine them into one logarithm, then rewrite in exponential form. Always check that every solution keeps each logarithm's argument positive.
Growth and decay models usually take the form \(N=N_0e^{kt}\), or \(N=N_0e^{-kt}\) for decay, where \(N_0\) is the starting amount, \(t\) is time and \(k\gt 0\) is a rate constant. For example, a drug in the bloodstream might follow \(C=80e^{-0.35t}\), where \(C\) is the concentration and \(t\) is in hours. After 3 hours, \(C=80e^{-1.05}=28.0\). The half-life is the time for the amount to halve. Setting \(e^{-0.35t}=\tfrac12\) and taking logs gives \(t=\dfrac{\ln 2}{0.35}=1.98\) hours. The half-life does not depend on the starting amount, which is a signature of exponential decay.
When both sides of an equation can be written as powers of the same base, you can equate the exponents directly. For instance, \(8^x=32\) becomes \(2^{3x}=2^5\), so \(3x=5\) and \(x=\tfrac53\). Use logarithms only when the bases cannot be matched.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(a^m\times a^n=a^{m+n}\), \(\;a^m\div a^n=a^{m-n}\), \(\;(a^m)^n=a^{mn}\) | Laws of exponents, used to simplify powers and to rewrite both sides with the same base. |
| \(a^{-n}=\dfrac{1}{a^n}\), \(\;a^{1/n}=\sqrt[n]{a}\) | Negative and fractional exponents. |
| \(\log_a x=y \iff a^y=x\) | Definition. Switch between log form and exponential form. |
| \(\log_a x+\log_a y=\log_a(xy)\) | Product law: adding logs means multiplying arguments. |
| \(\log_a x-\log_a y=\log_a\!\left(\dfrac{x}{y}\right)\) | Quotient law. |
| \(\log_a(x^m)=m\log_a x\) | Power law. Brings an exponent down to the front. |
| \(\log_a a=1\), \(\;\log_a 1=0\) | Two values worth knowing instantly. |
| \(\log_a x=\dfrac{\log_b x}{\log_b a}\) | Change of base. Use it to evaluate a logarithm in any base on a GDC. |
| \(a^x=b \Rightarrow x=\dfrac{\ln b}{\ln a}\) | Solving a simple exponential equation. |
| \(\ln x=\log_e x\), \(\;e^{\ln x}=x\) | Natural logarithm and its inverse. |
Worked examples
Example 1 Write \(2\ln 6-\ln 9+\ln 3\) as a single logarithm, then give its value to 3 s.f.
Use the power law on the first term: \(2\ln 6=\ln 6^2=\ln 36\).
The expression is now \(\ln 36-\ln 9+\ln 3\).
Subtracting logs divides, and adding logs multiplies: \(\ln\!\left(\dfrac{36\times 3}{9}\right)=\ln 12\).
Use the \(\ln\) key on your GDC: \(\ln 12=2.4849\ldots\)
Answer: \(\ln 12\approx 2.48\)
Example 2 Solve \(\log_2(x+1)+\log_2(x-1)=3\).
Combine the logarithms with the product law: \(\log_2\big[(x+1)(x-1)\big]=3\).
Expand the bracket: \(\log_2(x^2-1)=3\).
Write in exponential form: \(x^2-1=2^3=8\).
So \(x^2=9\) and \(x=3\) or \(x=-3\).
Check the arguments. If \(x=-3\) then \(x-1=-4\), and you cannot take the log of a negative number. If \(x=3\) then \(x+1=4\) and \(x-1=2\), which are both positive.
Answer: \(x=3\) only
Example 3 A savings account pays 4% interest per year, compounded annually. The balance after \(t\) years of an initial deposit of 2500 euros is \(B=2500(1.04)^t\). Find the number of whole years needed for the balance to exceed 4000 euros.
Set up the equation \(2500(1.04)^t=4000\).
Divide by 2500: \((1.04)^t=1.6\).
Take natural logarithms of both sides: \(t\ln 1.04=\ln 1.6\).
So \(t=\dfrac{\ln 1.6}{\ln 1.04}=\dfrac{0.47000}{0.039221}=11.98\ldots\)
Interest is added once a year, so \(t\) must be a whole number. At \(t=11\), \(B=3848.64\), which is too small. At \(t=12\), \(B=4002.58\), which is enough.
Answer: 12 years
Common errors
- Writing \(\log(x+y)=\log x+\log y\). The product law needs a product inside the log: \(\log x+\log y=\log(xy)\). There is no law for the log of a sum.
- Writing \(\dfrac{\log x}{\log y}=\log x-\log y\). The quotient law applies to \(\log x-\log y\) only. A fraction of two logs cannot be simplified this way, though it does equal \(\log_y x\) by change of base.
- Keeping a solution that makes the argument of a log zero or negative. Always substitute back and check.
- Dividing by \(\ln 1.04\) too early, or rounding \(\ln 1.04\) to 0.04. Keep full GDC values until the final line.
- Thinking \(\ln(x^2)\) and \((\ln x)^2\) are the same. The first equals \(2\ln x\). The second cannot be simplified.
Exam tip: When a question says "give an exact answer", leave the result as a quotient of logs or as a power of \(e\), such as \(\dfrac{\ln 50}{\ln 3}\). Only use the decimal when asked for 3 s.f.
Check yourself
- Evaluate \(\log_5 125+\log_5\!\left(\dfrac{1}{5}\right)\).
Answer
\(\log_5 125=3\) because \(5^3=125\), and \(\log_5\!\left(\dfrac15\right)=\log_5 5^{-1}=-1\). The sum is \(3+(-1)=2\). - Solve \(3^x=50\), giving your answer to 3 s.f.
Answer
Take logs: \(x\ln 3=\ln 50\), so \(x=\dfrac{\ln 50}{\ln 3}=\dfrac{3.9120}{1.0986}=3.56\). - Solve \(\ln(2x-1)=3\), giving your answer to 3 s.f.
Answer
Rewrite as \(2x-1=e^3=20.0855\ldots\), so \(2x=21.0855\ldots\) and \(x=10.5\). Check: \(2x-1\gt 0\), so the solution is valid. - For \(f(x)=2^x-3\), state the equation of the horizontal asymptote and the \(y\)-intercept.
Answer
As \(x\to-\infty\), \(2^x\to 0\), so \(f(x)\to -3\). The asymptote is \(y=-3\). At \(x=0\), \(f(0)=2^0-3=1-3=-2\), so the \(y\)-intercept is \((0,-2)\).