Exponential and Logarithmic Functions

Exponentials and logarithms are worth only a mark or two directly, but they are everywhere in Year 12: derivatives and integrals of $e^x$ and $\ln x$, growth and decay models, and financial maths. Fluency with the log laws saves a lot of time later.

• Exponential functions: graphs of $y=k{a}^x$ and $y=k{a}^{-x}$, asymptote, intercept, domain, range and behaviour as $x\to \pm \infty$.
• Euler's number $e\approx 2.71828$, and the result $\frac{d}{dx}{e}^x=e^x$.
• Logarithms: the definition, natural logs ($\ln x={\log }_{e}x$), the log laws and special values, change of base.
• Equations: solving $a^x=b$ and equations involving logs.
• Graphs: $y={\log }_{a}x$, and the fact that $y=a^x$ and $y={\log }_{a}x$ are reflections of each other in $y=x$.

For $y=a^x$ with $a>0$ and $a\ne 1$:

• The domain is all real $x$; the range is $y>0$.
• The $y$-intercept is $(0,1)$ and the horizontal asymptote is $y=0$.
• If $a>1$, the graph increases: $y\to \infty$ as $x\to \infty$ and $y\to 0$ as $x\to -\infty$. If $0<a<1$, it decreases.
• $y=a^{-x}$ is the reflection of $y=a^x$ in the $y$-axis, because $a^{-x}={\left(\frac{1}{a}\right)}^x$.

Multiplying by $k$ dilates the graph vertically: $y=k{a}^x$ has $y$-intercept $(0,k)$ and the same asymptote. If $k<0$, the graph is below the asymptote and the range is $y<0$.

Example. $y=3\times 2^{-x}$ has $y$-intercept 3, asymptote $y=0$, range $y>0$, and is decreasing. As $x\to \infty$, $y\to 0$; as $x\to -\infty$, $y\to \infty$.

Euler's number

For every base $a$, the gradient function of $a^x$ is a multiple of $a^x$: $\frac{d}{dx}(a^x)=k{a}^x$, where $k$ depends only on $a$. There is exactly one base for which $k=1$, so that the tangent at $(0,1)$ has gradient exactly 1. That base is Euler's number $e=2.71828\dots$, and it is the unique number for which

So $y=e^x$ is its own gradient function: at every point, the gradient equals the $y$-value.

${\log }_{a}y$ is the power you raise $a$ to in order to get $y$. For $a>0$, $a\ne 1$ and $y>0$:

So ${\log }_{2}8=3$ because $2^3=8$, and ${\log }_{10}0.01=-2$ because $10^{-2}=0.01$. You can only take the log of a positive number. The natural logarithm is $\ln x={\log }_{e}x$.

Because logs and exponentials undo each other:

The log laws

The log laws are the index laws in disguise. For example, $a^p{a}^q=a^{p+q}$ becomes the first law.

Change of base

In practice, use $b=e$ or $b=10$ to evaluate a log on a calculator: ${\log }_{2}50=\frac{\ln 50}{\ln 2}\approx 5.644$. It also converts between bases in algebra, for example ${\log }_{8}x=\frac{{\log }_{2}x}{{\log }_{2}8}=\frac{1}{3}{\log }_{2}x$.

Exponential equations

If you can write both sides as powers of the same base, equate the powers. Otherwise, take logs of both sides (or use the definition directly).

Worked example: an exponential equation

Solve $5^{x-1}=40$. Give the exact answer and the answer to 3 decimal places.

Check: $5^{2.292}\approx 40$. As a sense check, $5^2=25$ and $5^3=125$, so $x-1$ should be between 2 and 3.

An equation like $e^{2x}-7e^x+12=0$ is a quadratic in disguise: let $u=e^x$, solve $u^2-7u+12=0$, then solve $e^x=u$ for each value. Reject any $u\le 0$, since $e^x>0$.

Logarithmic equations

Use the log laws to combine everything into a single log on each side, then remove the logs. Always check every answer in the original equation: the log of a negative number or zero is undefined.

Worked example: a log equation with a rejected root

Solve $2{\log }_{6}x={\log }_6(x+12)$.

So $x=4$ or $x=-3$. In the original equation, ${\log }_{6}x$ needs $x>0$, so $x=-3$ is rejected. The solution is $x=4$. (Check: $2{\log }_{6}4={\log }_{6}16$ and ${\log }_6(4+12)={\log }_{6}16$.)

$y={\log }_{a}x$ is the inverse of $y=a^x$, so its graph is the reflection of $y=a^x$ in the line $y=x$. Swap the features:

For $a>1$, $y={\log }_{a}x$ increases slowly; for $0<a<1$ it decreases. The point $(a,1)$ always lies on $y={\log }_{a}x$.

• Writing $\log (m+n)=\log m+\log n$. There is no law for the log of a sum.
• Writing $\frac{\log m}{\log n}=\log m-\log n$. The quotient law is about the log of a quotient, not a quotient of logs.
• Writing ${\log }_a{m}^n=({\log }_{a}m{)}^n$. The power comes to the front as a multiplier.
• Not checking log equation answers in the original equation, or rejecting a valid negative answer for $x$ when only the argument of the log needs to be positive.
• Rounding $\ln 40$ and $\ln 5$ before dividing, which gives the wrong third decimal place.
• Confusing $y=a^{-x}$ (reflection in the $y$-axis) with $y=-a^x$ (reflection in the $x$-axis).

• Exact answers stay as logs, such as $1+{\log }_{5}40$ or $1+\frac{\ln 40}{\ln 5}$. Give a decimal only when asked.
• Show the log law you are using on each line. In "show that" questions, each correct application of a law is often what earns the marks.
• When you sketch an exponential or log graph, label the intercept and the asymptote and say what happens as $x\to \pm \infty$ if asked about behaviour.
• In growth and decay problems, write the equation, substitute, take logs, then give a sentence answer with units.

• Working with functions (Year 11): index laws and solving quadratics are used in nearly every equation here.
• Graph transformations (Year 11): translating and dilating exponential and log graphs, such as $y=2^{x-1}+3$.
• Differential and integral calculus (Year 12): derivatives and integrals of $e^x$, $a^x$ and $\ln x$.
• Sequences and series, Financial mathematics (Year 12): solving for the number of periods $n$ needs logarithms.
• Further graph transformations and modelling (Year 12): exponential models and logarithmic scales.