Exponentials and logarithms

Exponentials describe repeated multiplication and change at a rate proportional to the current amount. Logarithms answer the inverse question: what power produces this number? Together they provide the language for compound interest, population growth, radioactive decay and many scientific scales.

This overview connects the main ideas. Follow the linked lessons for fuller treatment of each technique.

Prerequisites

You should be able to:

• use positive, negative, fractional and zero indices
• rearrange equations and substitute into formulae
• recognise transformations of familiar graphs
• use a scientific calculator in exact and approximate form

1. Exponential functions

An exponential function has the variable in the exponent. Its basic form is

$$f(x)=a^x,$$

where

$$a>0 \qquad\text{and}\qquad a\ne1.$$

The restriction $a>0$ ensures $a^x$ is real for every real $x$. The value $a=1$ is excluded because $1^x=1$ is constant rather than genuine growth or decay.

The laws of indices give the central behaviour:

$$a^0=1,\qquad a^{-x}=\frac1{a^x},\qquad a^{x+y}=a^x a^y.$$

Therefore every graph $y=a^x$:

• passes through $(0,1)$
• has domain $\mathbb R$ and range $(0,\infty)$
• stays above the $x$ axis
• has horizontal asymptote $y=0$
• is increasing if $a>1$, and decreasing if $0<a<1$

For example:

$x$	$-2$	$-1$	$0$	$1$	$2$
$2^x$	$\frac14$	$\frac12$	$1$	$2$	$4$
$\left(\frac12\right)^x$	$4$	$2$	$1$	$\frac12$	$\frac14$

Notice that

$$\left(\frac12\right)^x=2^{-x}.$$

Its graph is the reflection of $y=2^x$ in the $y$ axis. Explore graph shape and transformations in exponential functions.

The special base $e$

The number

$$e=2.7182818\ldots$$

is the natural base for continuous change. The function $e^x$ has the remarkable property

$$\frac{\mathrm d}{\mathrm dx}e^x=e^x.$$

More generally,

$$\frac{\mathrm d}{\mathrm dx}a^x=a^x\ln a.$$

This is why models whose rate of change is proportional to their current value naturally take the form $y=Ae^{kx}$.

2. Logarithms undo exponentials

For $a>0$ and $a\ne1$,

$$\boxed{a^x=y\quad\Longleftrightarrow\quad x=\log_a y},$$

provided $y>0$.

Read $\log_a y$ as “the power of $a$ that gives $y$”.

Worked example 1: interpreting a logarithm

Evaluate $\log_3 81$.

Ask for the power $x$ satisfying

$$3^x=81.$$

Since $3^4=81$,

$$\boxed{\log_3 81=4}.$$

The same definition handles fractions and roots:

$$\log_2\left(\frac18\right)=-3 \quad\text{because}\quad 2^{-3}=\frac18,$$

and

$$\log_9 3=\frac12 \quad\text{because}\quad 9^{1/2}=3.$$

Two immediate results are

$$\log_a1=0,\qquad \log_a a=1.$$

On a calculator, $\log x$ usually means base $10$, while $\ln x$ means $\log_e x$. In A level work, $\ln$ is particularly important.

Logarithms need positive arguments

Over the real numbers, $\log_a x$ is defined only for $x>0$. There is no real value of $\ln 0$ or $\ln(-4)$. In an equation, every logarithm argument must be checked.

3. The logarithm laws

For positive $M$ and $N$,

$$\boxed{\log_a(MN)=\log_aM+\log_aN},$$ $$\boxed{\log_a\left(\frac MN\right)=\log_aM-\log_aN},$$

and, for real $k$,

$$\boxed{\log_a(M^k)=k\log_aM}.$$

These are consequences of the index laws, not unrelated rules to memorise. If $M=a^p$ and $N=a^q$, then

$$MN=a^{p+q},$$

so

$$\log_a(MN)=p+q=\log_aM+\log_aN.$$

Worked example 2: expanding a logarithm

Expand

$$\ln\left(\frac{x^3\sqrt y}{z^2}\right),$$

where $x,y,z>0$.

Apply the quotient law first:

$$\ln\left(\frac{x^3\sqrt y}{z^2}\right) =\ln(x^3\sqrt y)-\ln(z^2).$$

Then use the product and power laws:

$$\boxed{3\ln x+\frac12\ln y-2\ln z}.$$

Worked example 3: condensing logarithms

Write $2\log_a x-\log_a(x+1)$ as one logarithm.

Move the coefficient into a power:

$$2\log_a x=\log_a(x^2).$$

Then use the quotient law:

$$\boxed{\log_a\left(\frac{x^2}{x+1}\right)}.$$

The original expression requires $x>0$ and $x+1>0$, so its domain is $x>0$. See logarithms and their laws for proofs and further manipulation.

A common false law

There is no law that splits a sum:

$$\log_a(M+N)\ne\log_aM+\log_aN.$$

For instance, $\log_{10}(10+10)=\log_{10}20$, but $\log_{10}10+\log_{10}10=2$.

4. Inverse functions and graphs

The functions $y=a^x$ and $y=\log_a x$ are inverses. Therefore

$$\log_a(a^x)=x$$

for every real $x$, while

$$a^{\log_a x}=x$$

for $x>0$.

Their graphs are reflections of one another in the line $y=x$. Corresponding features exchange roles:

$y=a^x$	$y=\log_a x$
domain $\mathbb R$	range $\mathbb R$
range $(0,\infty)$	domain $(0,\infty)$
point $(0,1)$	point $(1,0)$
horizontal asymptote $y=0$	vertical asymptote $x=0$

This inverse viewpoint explains both why logarithms accept only positive inputs and why they can return any real output.

5. Solving exponential equations

Choose the simplest available method:

1. Rewrite both sides using a common base if possible.
2. Otherwise take logarithms of both sides.
3. If several related powers occur, use a substitution such as $u=2^x$.

Worked example 4: use a common base

Solve

$$8^{x-1}=4^{2x+1}.$$

Write both sides in base $2$:

$$(2^3)^{x-1}=(2^2)^{2x+1}.$$

Hence

$$2^{3x-3}=2^{4x+2}.$$

Equal powers of the same valid base have equal exponents:

$$3x-3=4x+2,$$

so

$$\boxed{x=-5}.$$

Worked example 5: take logarithms

Solve $5e^{0.3x}=17$.

First isolate the exponential:

$$e^{0.3x}=\frac{17}{5}.$$

Take natural logarithms:

$$\ln(e^{0.3x})=\ln\left(\frac{17}{5}\right).$$

Since $\ln(e^u)=u$,

$$0.3x=\ln\left(\frac{17}{5}\right),$$

so

$$\boxed{x=\frac{\ln(17/5)}{0.3}\approx4.08}.$$

Keep the exact logarithmic form until the final line to avoid premature rounding.

Worked example 6: substitution

Solve

$$4^x-5(2^x)+4=0.$$

Since $4^x=(2^x)^2$, let $u=2^x$. Then $u>0$ and

$$u^2-5u+4=0.$$

Factorise:

$$(u-1)(u-4)=0.$$

Thus $u=1$ or $u=4$. Returning to $2^x$ gives

$$2^x=1\Rightarrow x=0, \qquad 2^x=4\Rightarrow x=2.$$

Therefore

$$\boxed{x=0\text{ or }x=2}.$$

For logarithmic equations, domain checks and equations without closed forms, continue to exponential and logarithmic equations.

6. Exponential growth and decay

A quantity that changes by the same proportion in equal time intervals can be modelled discretely by

$$Q=Q_0b^t,$$

where $Q_0$ is the initial value. Growth has $b>1$ and decay has $0<b<1$.

Continuous proportional change is modelled by

$$\boxed{Q=Q_0e^{kt}}.$$

Here $k>0$ gives growth and $k<0$ gives decay. The parameter $k$ has units of inverse time.

Worked example 7: half life

A substance has mass $80$ g initially and half life $6$ hours. Find its mass after $15$ hours.

Each $6$ hour period multiplies the mass by $\frac12$, so

$$M=80\left(\frac12\right)^{t/6}.$$

At $t=15$,

$$M=80\left(\frac12\right)^{15/6} =80\left(\frac12\right)^{5/2} =10\sqrt2.$$

Therefore

$$\boxed{M\approx14.1\text{ g}}.$$

The exponent $15/6$ counts how many half life periods have elapsed. It need not be an integer.

Worked example 8: find a growth constant

A population follows $P=1200e^{kt}$ and reaches $1800$ after $4$ years. Find $k$.

Substitute $t=4$ and $P=1800$:

$$1800=1200e^{4k}.$$

Divide by $1200$:

$$e^{4k}=\frac32.$$

Take natural logarithms:

$$4k=\ln\left(\frac32\right),$$

so

$$\boxed{k=\frac14\ln\left(\frac32\right)\approx0.1014\text{ year}^{-1}}.$$

The model and its assumptions are developed in exponential growth and decay.

7. Linearising relationships

Taking logarithms can turn a curved relationship into a straight line.

If

$$y=Ae^{kx},$$

then

$$\ln y=\ln A+kx.$$

Thus a plot of $\ln y$ against $x$ has gradient $k$ and intercept $\ln A$.

For a power law $y=Ax^n$,

$$\ln y=\ln A+n\ln x,$$

so a plot of $\ln y$ against $\ln x$ has gradient $n$. The choice of axes matters: plotting the wrong transformed variables will not reveal the intended straight line. Learn how to extract parameters and assess models in linearising exponential and power relationships.

Misconceptions to avoid

• $a^x$ is not $ax$. In $a^x$, the variable controls repeated multiplication.
• Exponential growth means a constant percentage change, not a constant amount added.
• $\ln(x^2)=2\ln x$ is valid only where the original and rewritten real expressions have matching domains. For arbitrary $x\ne0$, $\ln(x^2)=2\ln|x|$.
• Taking logarithms does not mean taking the logarithm of separate terms: $\ln(u+v)$ cannot be split.
• A negative decay constant does not make the quantity negative. If $Q_0>0$, then $Q_0e^{kt}>0$ for every real $t$.
• Calculator values should be rounded only at the end of a calculation.

Self check

1. Evaluate $\log_5(1/125)$.
2. Simplify $\ln 12+\ln 3-\ln 2$.
3. Solve $3^{2x-1}=7$, giving an exact answer.
4. Solve $9^x-10(3^x)+9=0$.
5. A quantity follows $Q=500e^{-0.08t}$. State its initial value and whether it grows or decays.
6. If $y=7x^{2.4}$, what are the gradient and intercept of a graph of $\ln y$ against $\ln x$?

Answers

1. $-3$, because $5^{-3}=1/125$.
2. $\ln(12\times3/2)=\boxed{\ln18}$.
3. Taking logarithms gives $(2x-1)\ln3=\ln7$, so $\boxed{x=\frac12\left(1+\frac{\ln7}{\ln3}\right)}$.
4. Let $u=3^x>0$. Then $u^2-10u+9=(u-1)(u-9)=0$, giving $\boxed{x=0\text{ or }x=2}$.
5. $\boxed{Q_0=500}$. It decays because the coefficient of $t$ in the exponent is negative.
6. Since $\ln y=\ln7+2.4\ln x$, the gradient is $\boxed{2.4}$ and the intercept is $\boxed{\ln7}$.

What to study next

Work through the topic in this order:

1. Exponential functions for graphs, transformations and the base $e$.
2. Logarithms and their laws for inverse functions and algebraic manipulation.
3. Exponential and logarithmic equations for exact, substituted and numerical solutions.
4. Exponential growth and decay for modelling, half life and doubling time.
5. Linearising exponential and power relationships for transformed data and parameter estimation.

These ideas lead directly into calculus, where exponential and logarithmic functions are differentiated, integrated and used in differential equations.