Notes · MathematicsUK · A-Levels

Exponentials and logarithms

This topic introduces exponential functions, the special number e and the natural logarithm, the laws of logarithms, the solution of exponential equations using logs, and the use of exponential functions to model growth and decay. A key applied skill is linearising a power or exponential relationship so that its parameters can be estimated from a straight-line log graph.

4 sections~10 min reading time3 competenciesLevel Standard 3 · Advanced 1

T·0777 / 20

Exam profile

AO1 · Use the laws of logarithms and solve exponential equationsAO2 · Justify the log-linearising method and interpret log graphsAO3 · Model growth and decay and estimate parameters from data

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basic level

AS-Level covers $y = a^x$ , the laws of logarithms and solving equations of the form $a^x = b$ .

higher level

The full A-Level adds $e^x$ and its derivative, natural logarithms, exponential models and the use of log graphs to estimate parameters.

Depth

Text

Contents · 4 sections

Exponentials and logarithms

Exponential functions and the number e#

StandardDfE GCE Mathematics F1-F2AQA 7357 F

e to the x and its inverse ln x

Fig. 1

y = e^x

and

y = \ln x

are reflections in

y = x

; one passes through

(0,1)

, the other through

(1,0)

An exponential function

y = a^x

(with

a > 0

a \neq 1

) has the variable in the exponent, so it multiplies by a fixed factor for each unit increase in

x

. Its graph passes through

(0, 1)

, is always positive, and is increasing when

a > 1

and decreasing when

0 < a < 1

; the

x

-axis is a horizontal asymptote. This constant-ratio behaviour is what makes exponentials the natural model for growth and decay.

Among all possible bases, one is special: the number

e \approx 2.718

is the base for which the gradient of

y = a^x

at every point equals the value of the function itself. That is,

\dfrac{d}{dx}e^x = e^x

. This self-derivative property is why

e

is called the natural base and why

e^x

is central to calculus; more generally

\dfrac{d}{dx}e^{kx} = ke^{kx}

The natural logarithm

\ln x = \log_e x

is the inverse of

e^x

, so

\ln(e^x) = x

and

e^{\ln x} = x

for

x > 0

. Consequently the graph of

y = \ln x

is the reflection of

y = e^x

in the line

y = x

: it passes through

(1, 0)

, increases without bound but ever more slowly, and has the

y

-axis as a vertical asymptote (it is defined only for

x > 0

Recognising the pairing of

e^x

and

\ln x

as mutual inverses is the practical key to the topic: taking natural logs undoes an exponential, and exponentiating undoes a logarithm. This is exactly the move that solves equations with the unknown in the exponent, and it turns exponential models into linear ones that can be analysed with a straight line.

\frac{d}{dx}e^{kx} = k\,e^{kx}

Derivative of the exponential

The defining property of $e$ : the exponential is its own derivative (up to the constant $k$ ).

\ln(e^x) = x, \qquad e^{\ln x} = x\ (x > 0)

e and ln are inverse

Each function undoes the other; this is how exponential equations are solved.

Worked example

Using inverse functions

Solve $e^{2x} = 7$ , giving your answer to 3 significant figures.

Take natural logs
$\ln(e^{2x}) = \ln 7 \Rightarrow 2x = \ln 7$ .
Solve for x
$x = \dfrac{\ln 7}{2} = \dfrac{1.9459}{2}$ .
$x = \frac{\ln 7}{2}$
Evaluate
$x = 0.973$ (3 s.f.).

Result: $x = \dfrac{\ln 7}{2} \approx 0.973$ .

Typical mistakes

Treating $e$ as if it were a variable rather than the fixed constant $\approx 2.718$ .
Trying to take $\ln$ of a non-positive number, which is undefined.

Active revision

Sketch $y = e^{-x}$ and $y = \ln(x + 2)$ on the same axes, marking all intercepts and asymptotes.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Logarithms and the laws of logs#

StandardDfE GCE Mathematics F3-F4AQA 7357 F

The natural logarithm graph

Fig. 2

y = \ln x

passes through

(1, 0)

and

(e, 1)

and is defined only for

x > 0

A logarithm answers the question 'to what power must the base be raised?':

\log_a x = y

means exactly

a^y = x

. Logarithms and exponentials are therefore two descriptions of the same relationship, and every log statement can be rewritten as an index statement and vice versa. This equivalence,

\log_a x = y \Leftrightarrow a^y = x

, is the definition to fall back on whenever you are unsure.

The laws of logarithms convert products, quotients and powers into sums, differences and multiples:

\log_a(xy) = \log_a x + \log_a y

\log_a\!\left(\dfrac{x}{y}\right) = \log_a x - \log_a y

, and

\log_a(x^n) = n\log_a x

. They follow directly from the index laws, because logs are exponents. The special cases

\log_a a = 1

and

\log_a 1 = 0

are used constantly.

The power law is the single most useful, because it brings an unknown exponent down to ground level where it can be solved for. It is exactly why taking logs of both sides of

a^x = b

gives

x\log a = \log b

and hence

x = \dfrac{\log b}{\log a}

. Logs may be taken to any base for this, but base

10

(written

\log

) and base

e

(written

\ln

) are the two on the calculator.

Care with the laws is essential: there is no rule for

\log_a(x + y)

, which is not

\log_a x + \log_a y

, and

\dfrac{\log x}{\log y}

is not

\log\dfrac{x}{y}

. The laws apply to the argument of a single log, not to logs added or divided outside. Keeping this straight prevents the commonest algebraic errors in the whole topic.

\log_a(xy) = \log_a x + \log_a y, \quad \log_a\!\left(\tfrac{x}{y}\right) = \log_a x - \log_a y

Product and quotient laws

Products become sums and quotients become differences of logarithms.

\log_a(x^n) = n\log_a x

Power law

Brings an exponent down as a multiplier — the key to solving for an unknown power.

Worked example

Combining logarithms

Solve $\log_3(2x + 1) - \log_3(x - 1) = 1$ .

Combine using the quotient law
$\log_3\!\left(\dfrac{2x + 1}{x - 1}\right) = 1$ .
Rewrite as an index equation
$\dfrac{2x + 1}{x - 1} = 3^1 = 3$ .
$\frac{2x + 1}{x - 1} = 3$
Solve
$2x + 1 = 3(x - 1) \Rightarrow 2x + 1 = 3x - 3 \Rightarrow x = 4$ .
Check validity
$x = 4$ gives positive arguments $2x+1 = 9$ and $x - 1 = 3$ , so it is valid.

Result: $x = 4$ .

Typical mistakes

Writing $\log(x + y) = \log x + \log y$ , which is false.
Confusing $\dfrac{\log b}{\log a}$ with $\log\dfrac{b}{a}$ .

Active revision

Solve $\log_2 x + \log_2(x - 2) = 3$ , checking that your solution is valid.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Solving exponential equations#

StandardDfE GCE Mathematics F5AQA 7357 F

Exponential decay and half-life

Fig. 3

N = 100e^{-0.2t}

falls to half its value when

t = \dfrac{\ln 2}{0.2} \approx 3.47

The standard method for an equation with the unknown in the exponent is to take logarithms of both sides and use the power law. For

a^x = b

this gives

x\log a = \log b

, so

x = \dfrac{\log b}{\log a}

; either base of logarithm works, provided you use the same base on both sides. This is the direct application of logarithms as 'exponent-finders'.

Some exponential equations are secretly quadratics. An equation such as

e^{2x} - 5e^{x} + 6 = 0

becomes, with the substitution

u = e^{x}

, the quadratic

u^2 - 5u + 6 = 0

; solving gives

u = 2

u = 3

, and then

x = \ln 2

x = \ln 3

. Spotting the hidden quadratic — usually signalled by a term in

a^{2x}

and a term in

a^{x}

— is a common exam demand.

When the exponentials have different bases you take logs and keep the structure. For

2^{x} = 3^{x - 1}

, taking logs gives

x\log 2 = (x - 1)\log 3

; expanding and collecting the

x

terms leads to a linear equation in

x

that you solve algebraically. The key is to treat

\log 2

and

\log 3

as ordinary constants.

Whichever route you take, keep exact forms (

\ln 2

\dfrac{\log 5}{\log 3}

) until the final step, then round only if a decimal is requested. Always check that any solution is consistent with the original equation, discarding, for instance, a value of

u = e^x

that is negative, since an exponential can never be negative.

a^x = b \ \Rightarrow\ x = \frac{\log b}{\log a}

Solving for an unknown exponent

Take logs of both sides and divide; any consistent base of log may be used.

Worked example

A hidden quadratic

Solve $e^{2x} - 7e^{x} + 12 = 0$ , giving answers in terms of natural logarithms.

Substitute
Let $u = e^{x}$ ; then $u^2 - 7u + 12 = 0$ .
Factorise
$(u - 3)(u - 4) = 0$ , so $u = 3$ or $u = 4$ .
Return to x
$e^x = 3 \Rightarrow x = \ln 3$ ; $e^x = 4 \Rightarrow x = \ln 4$ . Both are positive, so both are valid.

Result: $x = \ln 3$ or $x = \ln 4$ .

Typical mistakes

Accepting a negative value of $u = e^x$ as a solution (impossible, since $e^x > 0$ ).
Dividing by $\log a$ incorrectly, or mixing logs of different bases across the equation.

Active revision

Solve $3^{2x} - 10\times 3^{x} + 9 = 0$ , giving exact answers.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Modelling with exponentials and log-linear graphs#

AdvancedDfE GCE Mathematics F6-F7AQA 7357 F

A log-linear plot of exponential data

Fig. 4For

y = Ab^x

, the points

(x, \log y)

lie on a straight line; its gradient is

\log b

and its intercept is

\log A

Exponential functions model any quantity with a constant proportional rate of change: population growth, radioactive decay, cooling and compound interest. A growth model has the form

y = Ae^{kt}

with

k > 0

; a decay model has

k < 0

. The parameters

A

(the initial value) and

k

(the rate) carry the meaning, and interpreting them in context — 'the population doubles every 12 years' — is an examined skill.

The great practical difficulty is estimating

A

and

k

from data, and the solution is to linearise. Taking logs of

y = Ae^{kt}

gives

\ln y = \ln A + kt

, which is linear in

t

: plotting

\ln y

against

t

yields a straight line of gradient

k

and intercept

\ln A

. The awkward exponential becomes a line whose gradient and intercept can be read off.

The other common model is a power law

y = Ax^n

, which linearises differently: taking logs gives

\log y = \log A + n\log x

, linear in

\log x

. So you plot

\log y

against

\log x

for a power law, but

\ln y

(or

\log y

) against

t

for an exponential. Choosing the correct pair of axes is itself a decision the question tests, and it is settled by which variable appears in the exponent.

Once the line is fitted (often by a regression line through the transformed data), you convert back: the gradient gives

n

k

directly, and the intercept gives

\log A

\ln A

, from which

A

follows by exponentiating. The final model is then used to predict, always with an eye on the range over which the data — and hence the model — can be trusted.

y = Ae^{kt} \ \Rightarrow\ \ln y = \ln A + kt

Linearising an exponential model

Plot $\ln y$ against $t$ : gradient $k$ , intercept $\ln A$ .

y = Ax^{n} \ \Rightarrow\ \log y = \log A + n\log x

Linearising a power law

Plot $\log y$ against $\log x$ : gradient $n$ , intercept $\log A$ .

Worked example

From a log-linear line to the model

Values of $y = Ae^{kt}$ are plotted as $\ln y$ against $t$ , giving a straight line through $(0, 2)$ with gradient $0.5$ . Find $A$ and $k$ .

Match to the linear form
$\ln y = \ln A + kt$ , so the gradient is $k$ and the intercept is $\ln A$ .
Read the gradient
Gradient $= 0.5$ , so $k = 0.5$ .
Read the intercept
Intercept $= 2 = \ln A$ , so $A = e^{2} = 7.39$ (3 s.f.).
$A = e^{2} \approx 7.39$

Result: $k = 0.5$ and $A = e^{2} \approx 7.39$ , so $y \approx 7.39\,e^{0.5t}$ .

Typical mistakes

Plotting $\log y$ against $x$ for a power law (it should be against $\log x$ ).
Forgetting to exponentiate the intercept to recover $A$ from $\ln A$ or $\log A$ .

Active revision

Data for $y = Ab^{x}$ gives a straight line of $\ln y$ against $x$ with gradient $0.35$ and intercept $1.6$ . Find $A$ and $b$ , and predict $y$ when $x = 10$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Exponentials and logarithms

Key points

Using inverse functions

Key points

Combining logarithms

Key points

A hidden quadratic

Key points

From a log-linear line to the model

Sources