Exponentials and Logarithms Overview

This topic is DfE content area F · Exponentials and logarithms. It is about one pair of operations that undo each other: raising a number to a power, and asking what power was used. The first gives exponential functions such as $2^x$ and $e^x$; the second gives logarithms.

At GCSE you met exponential growth as repeated multiplication — compound interest, depreciation. At A Level you can solve for the unknown time ("when will it reach …?") because logarithms let you bring an unknown down out of an index. You also meet $e$, the base whose exponential function is its own gradient.

What is new compared with GCSE

The function $e^x$ and its gradient · $\ln x$ as the inverse of $e^x$
Logarithms and their laws · solving $a^x = b$
Fitting models by plotting $\log y$ against $x$ or $\log x$

Calculators are allowed in every A Level paper, but many log questions say "give your answer in exact form" or "show detailed reasoning". Then you must write the method and leave answers such as $\ln 5$ or $\dfrac{\ln 2}{\ln 3 - \ln 2}$, not a decimal from an equation solver.

Word	Meaning	Example
Base	The number being raised to a power.	The $2$ in $2^x$ or in $\log_2 x$
Index / exponent	The power.	The $x$ in $2^x$
Exponential function	A function with the variable in the index.	$y = 3^x$, $y = 5e^{-0.2x}$
$e$	The constant $2.71828\ldots$ for which $e^x$ has gradient $e^x$.	$\dfrac{\mathrm{d}}{\mathrm{d}x}e^x = e^x$
Logarithm	The power you raise the base to.	$\log_2 8 = 3$
Natural logarithm	Logarithm to base $e$.	$\ln x = \log_e x$
Inverse	A function that undoes another.	$\ln\left(e^x\right) = x$
Asymptote	A line the curve approaches but never meets.	$y = 0$ for $y = e^x$
Growth / decay	Increasing / decreasing exponentially.	$Ae^{kt}$ with $k > 0$ / $k < 0$

1 — Switching between forms

$\log_a b = c \iff a^c = b$

2 — The laws of logarithms

$\log xy = \log x + \log y \qquad \log\dfrac{x}{y} = \log x - \log y \qquad \log x^k = k\log x$

3 — Solving $a^x = b$

Take logs: $x\ln a = \ln b$, so $x = \dfrac{\ln b}{\ln a}$

4 — The gradient of an exponential

$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(Ae^{kx}\right) = Ake^{kx}$ — the rate of change is proportional to the value

5 — Reduction to linear form

$y = ab^x \Rightarrow \log y = \log a + x\log b$ $y = ax^n \Rightarrow \log y = \log a + n\log x$

Worked Example — Several techniques in one question

Solve $3^{2x} - 3^{x+2} + 14 = 0$, giving your answers to 3 significant figures.

①Rewrite the terms: $3^{2x} = \left(3^x\right)^2$ and $3^{x+2} = 9 \times 3^x$.

②Let $y = 3^x$: $y^2 - 9y + 14 = 0$.

③$(y - 2)(y - 7) = 0$, so $3^x = 2$ or $3^x = 7$.

④Take logs: $x = \dfrac{\ln 2}{\ln 3} = 0.631$ or $x = \dfrac{\ln 7}{\ln 3} = 1.77$.

Index laws, a hidden quadratic, and logs to finish — the typical shape of an exam question on this topic.

Content area F, split into four pages.

FExponential Functions and eGraphs of $y = a^x$, the number $e$, the gradient of $e^{kx}$, and $y = \ln x$ as the inverse.
FLogarithms and Their LawsThe definition, the three laws, natural logs, and switching between index and log form.
FExponential and Log EquationsSolving $a^x = b$, hidden quadratics in $e^x$ or $2^x$, and equations involving logs.
FExponential ModellingGrowth and decay, fitting straight lines to logged data, and the limitations of models.

Inventing a law for $\log(x + y)$. There isn't one. $\ln(x + 3) \neq \ln x + \ln 3$.

Dividing logs instead of subtracting. $\dfrac{\ln 8}{\ln 2} = 3$, but $\ln 8 - \ln 2 = \ln 4$. These are different things.

Keeping impossible roots. $e^x = -2$ and $\ln(-4)$ have no meaning. State that you reject them and why.

Forgetting to flip an inequality. $\ln 0.8 < 0$. Dividing by it reverses the sign.

Misreading the intercept. On a $\log y$ against $\log x$ graph the intercept is $\log a$, found where $x = 1$, and $a$ itself is $10^{\text{intercept}}$ (or $e^{\text{intercept}}$ for $\ln$).

$y = a^x$

Through $(0,1)$; asymptote $y = 0$.

$y = \ln x$

Through $(1,0)$; asymptote $x = 0$; only $x > 0$.

Inverses

$e^{\ln x} = x$, $\ln e^x = x$

Definition

$\log_a b = c \iff a^c = b$

Laws

Multiply → add; divide → subtract; power → multiply.

$a^x = b$

$x = \ln b / \ln a$

Gradient

$\dfrac{\mathrm{d}}{\mathrm{d}x}e^{kx} = ke^{kx}$

Models

$N = Ae^{kt}$, $\dfrac{\mathrm{d}N}{\mathrm{d}t} = kN$

Linear form

$\log y$ v $x$ for $ab^x$; $\log y$ v $\log x$ for $ax^n$.

Ten questions drawing on the whole of content area F. Work each one through before opening the solution.

Question 1

Without a calculator, find $\log_2 64 + \log_3 \tfrac19$.

▶ Show solution

$2^6 = 64$, so $\log_2 64 = 6$. $3^{-2} = \tfrac19$, so $\log_3\tfrac19 = -2$.

Total: $6 + (-2) = 4$.

Question 2

Solve $6^x = 15$, giving your answer to 3 significant figures.

▶ Show solution

$x = \dfrac{\ln 15}{\ln 6} = 1.51$ (3 s.f.)

Question 3

Show that $\ln 12 - \ln 3 + 2\ln\tfrac12 = 0$.

▶ Show solution

$2\ln\tfrac12 = \ln\tfrac14$.

$\ln\left(\dfrac{12}{3} \times \dfrac14\right) = \ln 1 = 0$.

Question 4

Find the gradient of $y = 7e^{-3x}$ at the point where $x = 0$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = -21e^{-3x}$. At $x = 0$: $-21$.

Question 5

Sketch $y = e^x - 4$, giving the equation of the asymptote and the exact coordinates of the axis intercepts.

▶ Show solution

$y = e^x$ translated down $4$: asymptote $y = -4$ (approached as $x \to -\infty$).

$x = 0$: $y = 1 - 4 = -3$, so $(0, -3)$.

$y = 0$: $e^x = 4 \Rightarrow x = \ln 4$, so $(\ln 4,\, 0)$.

Question 6

Solve $e^{2x} - 3e^x - 10 = 0$, giving an exact answer.

▶ Show solution

Let $y = e^x$: $y^2 - 3y - 10 = 0 \Rightarrow (y - 5)(y + 2) = 0$.

$e^x = -2$ is impossible as $e^x > 0$.

$e^x = 5 \Rightarrow x = \ln 5$.

Question 7

Solve $\log_4 x + \log_4(x - 6) = 2$.

▶ Show solution

$\log_4\left(x(x-6)\right) = 2 \Rightarrow x^2 - 6x = 16 \Rightarrow x^2 - 6x - 16 = 0$.

$(x - 8)(x + 2) = 0$. Reject $x = -2$ (log of a negative).

$x = 8$. Check: $\log_4 8 + \log_4 2 = \log_4 16 = 2$ ✓

Question 8

Data are modelled by $y = ab^x$. A graph of $\log_{10} y$ against $x$ is a straight line with equation $\log_{10} y = 1.3 - 0.1x$. Find $a$ and $b$, and say whether $y$ is growing or decaying.

▶ Show solution

$\log_{10} a = 1.3 \Rightarrow a = 10^{1.3} \approx 20.0$.

$\log_{10} b = -0.1 \Rightarrow b = 10^{-0.1} \approx 0.794$.

$b < 1$, so $y$ is decaying (the line has negative gradient).

Question 9

(a) Show that $\ln\left(x^2 - 1\right) - \ln(x + 1) = \ln(x - 1)$ for $x > 1$. (b) Hence solve $\ln\left(x^2 - 1\right) - \ln(x + 1) = 2$, giving an exact answer.

▶ Show solution

(a) $\ln\left(\dfrac{x^2 - 1}{x + 1}\right) = \ln\left(\dfrac{(x-1)(x+1)}{x+1}\right) = \ln(x - 1)$.

(b) $\ln(x - 1) = 2 \Rightarrow x - 1 = e^2 \Rightarrow x = 1 + e^2$ $(\approx 8.39)$.

Question 10

The concentration of a drug in a patient's blood $t$ hours after a dose is modelled by $C = 12e^{-0.25t}$ mg/l.

(a) State the initial concentration. (b) The drug is ineffective below $3$ mg/l. Find, in exact form, how long the dose remains effective. (c) Find the rate at which the concentration is falling when $t = 2$. (d) A second, identical dose is given at $t = 4$. Assuming the effects add, find the concentration immediately afterwards.

▶ Show solution

(a) $12$ mg/l.

(b) $12e^{-0.25t} = 3 \Rightarrow e^{-0.25t} = \tfrac14 \Rightarrow -0.25t = -\ln 4 \Rightarrow t = 4\ln 4$ hours $(\approx 5.55)$.

(c) $\dfrac{\mathrm{d}C}{\mathrm{d}t} = -3e^{-0.25t}$. At $t = 2$: $-3e^{-0.5} \approx -1.82$. Falling at $1.82$ mg/l per hour.

(d) Remaining from the first dose: $12e^{-1} \approx 4.41$. Total $\approx 4.41 + 12 = 16.4$ mg/l.