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Cambridge International AS & A Level Mathematics 9709 · Pure Mathematics 3

Logarithmic and exponential functions

Logarithms as indices and the laws of logarithms; the functions $e^{x}$ and $\ln x$ , their graphs and their relationship as inverses; solving equations and inequalities with the unknown in an index; and using logarithms to reduce $y = k x^{n}$ and $y = k (a^{x})$ to straight lines.

3.2Paper 3: Pure Mathematics 3Free lesson

What the syllabus says

3.2 (Paper 3)

understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base)
understand the definition and properties of $e^{x}$ and $\ln x$ , including their relationship as inverse functions and their graphs
use logarithms to solve equations and inequalities in which the unknown appears in indices
use logarithms to transform a given relationship to linear form, and hence determine unknown constants by considering the gradient and/or intercept.

Notes and examples

Including knowledge of the graph of $y = e^{k x}$ for both positive and negative values of $k$ .
e.g. $2^{x} < 5$ , $3 \times 2^{3 x - 1} < 5$ , $3^{x + 1} = 4^{2 x - 1}$ .
e.g. $y = k x^{n}$ gives $\ln y = \ln k + n \ln x$ which is linear in $\ln x$ and $\ln y$ . $y = k (a^{x})$ gives $\ln y = \ln k + x \ln a$ which is linear in $x$ and $\ln y$ .

Quoted from the Cambridge International AS and A Level Mathematics 9709 syllabus (2026-2027 and 2028-2030 editions, which share this content).

Logarithms and indices Paper 3

A logarithm is an index. The statement
$\log_{a} x = y ⟺ a^{y} = x (a > 0, a \neq 1, x > 0)$
says that $\log_{a} x$ is the power to which $a$ must be raised to give $x$ . So $\log_{2} 32 = 5$ because $2^{5} = 32$ , and $\log_{10} 0.01 = - 2$ because $10^{- 2} = 0.01$ . Since $a^{y}$ is always positive, the logarithm of zero or of a negative number is not defined.

Two values follow at once: $\log_{a} a = 1$ (as $a^{1} = a$ ) and $\log_{a} 1 = 0$ (as $a^{0} = 1$ ).

The laws of logarithms are the laws of indices rewritten. If $x = a^{m}$ and $y = a^{n}$ then $x y = a^{m + n}$ , so the logarithm of a product is the sum of the logarithms:

$\log_{a} (x y) = \log_{a} x + \log_{a} y$
$\log_{a} (\frac{x}{y}) = \log_{a} x - \log_{a} y$
$\log_{a} (x^{k}) = k \log_{a} x$ , for any real $k$ , including $\log_{a} \frac{1}{x} = - \log_{a} x$ and $\log_{a} \sqrt{x} = \frac{1}{2} \log_{a} x$ .

There is no law for $\log_{a} (x + y)$ , and $\frac{\log_{a} x}{\log_{a} y}$ is not $\log_{a} \frac{x}{y}$ . Changing the base of a logarithm is not needed on this paper.

Solving logarithmic equations. Use the laws to collect each side into a single logarithm, then remove the logarithm by writing the index form. Finally check each answer in the original equation: any value that makes the argument of a logarithm zero or negative must be rejected.

Worked example 1 Paper 3

Solve the equation $2 \log_{3} x - \log_{3} (x - 2) = 2$ .

Power law, then quotient law: $\log_{3} x^{2} - \log_{3} (x - 2) = \log_{3} \frac{x^{2}}{x - 2}$ .
So $\log_{3} \frac{x^{2}}{x - 2} = 2$ , which means $\frac{x^{2}}{x - 2} = 3^{2} = 9$ .
$x^{2} = 9 x - 18$ , so $x^{2} - 9 x + 18 = 0$ and $(x - 3) (x - 6) = 0$ .
Check: both $x = 3$ and $x = 6$ make $x$ and $x - 2$ positive, so both are valid.

Answer: $x = 3$ or $x = 6$

Check yourself Paper 3

Solve the equation $\log_{2} (x + 5) - \log_{2} (x - 1) = 3$ .

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The exponential function and the natural logarithm Paper 3

The number $e = 2.71828 \dots$ is the base for which the curve $y = e^{x}$ has gradient exactly 1 where it crosses the $y$ -axis; in fact its gradient at every point equals its height. The exponential function $e^{x}$ is defined for every real $x$ , and:

$e^{x} > 0$ for every $x$ , so the graph lies above the $x$ -axis;
it is increasing, passes through $(0, 1)$ , and approaches 0 as $x \to - \infty$ , so $y = 0$ is an asymptote.

The natural logarithm is the logarithm to base $e$ : $\ln x = \log_{e} x$ . It obeys all the laws above, and it is the inverse of $e^{x}$ :
$\ln (e^{x}) = x for every x, e^{\ln x} = x for x > 0$
Because they are inverse functions, the graph of $y = \ln x$ is the reflection of $y = e^{x}$ in the line $y = x$ . So $y = \ln x$ is defined only for $x > 0$ , passes through $(1, 0)$ , increases slowly without limit, and has the $y$ -axis as an asymptote. Every point $(a, b)$ on one curve matches the point $(b, a)$ on the other.

The graph of $y = e^{k x}$ . Every such curve passes through $(0, 1)$ and has $y = 0$ as an asymptote.

For $k > 0$ it is increasing: exponential growth, steeper for larger $k$ .
For $k < 0$ it is decreasing: exponential decay. $y = e^{- x}$ is the reflection of $y = e^{x}$ in the $y$ -axis.

A curve like $y = 3 e^{- 2 x} + 1$ is $y = e^{- 2 x}$ stretched by factor 3 parallel to the $y$ -axis and translated up 1: it crosses the $y$ -axis at 4 and has asymptote $y = 1$ .

$y = e^{x}$ and $y = \ln x$ are reflections of each other in $y = x$ .

Positive $k$ gives growth; negative $k$ gives decay. Both pass through $(0, 1)$ .

Worked example 2 Paper 3

(a) Solve the equation $e^{2 x} - 5 e^{x} + 6 = 0$ , giving exact answers.
(b) Solve the equation $\ln (x + 1) = 2$ , giving an exact answer.

(a) $e^{2 x} = (e^{x})^{2}$ , so with $u = e^{x}$ this is $u^{2} - 5 u + 6 = 0$ .
$(u - 2) (u - 3) = 0$ , so $e^{x} = 2$ or $e^{x} = 3$ .
Take natural logarithms: $x = \ln 2$ or $x = \ln 3$ .
(b) Apply the inverse, $e$ to the power of each side: $x + 1 = e^{2}$ , so $x = e^{2} - 1$ .

Answer: (a) $x = \ln 2$ or $x = \ln 3$ (b) $x = e^{2} - 1$

Check yourself Paper 3

Solve the equation $\ln (3 x) = 2$ , giving an exact answer.

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Equations and inequalities with the unknown in an index Paper 3

When the unknown is in an index, take logarithms of both sides, using any base (usually $\ln$ , or $\log_{10}$ ). The power law brings the index down:
$3^{x + 1} = 4^{2 x - 1}$ gives
$(x + 1) \ln 3 = (2 x - 1) \ln 4$
which is linear in $x$ . Collect the $x$ terms and divide. Keep logarithms exact until the last line, then use a calculator.

When the equation has terms such as $2^{2 x}$ and $2^{x}$ , it is a quadratic in disguise: put $u = 2^{x}$ , so $2^{2 x} = u^{2}$ and $2^{x + 1} = 2 u$ . Solve for $u$ , and remember that $u = 2^{x}$ must be positive, so a negative root gives no solution.

Inequalities. $\ln$ is an increasing function, so taking logarithms of both sides of an inequality keeps its direction: $2^{x} < 5$ gives $x \ln 2 < \ln 5$ , and since $\ln 2 > 0$ , $x < \frac{\ln 5}{\ln 2} = 2.32$ . The danger comes next: dividing by a negative logarithm reverses the inequality. The logarithm of any number between 0 and 1 is negative, so this happens with ${0.8}^{n}$ , $\frac{1}{2}$ , or $\ln 2 - \ln 3$ . Such inequalities often appear in context, asking for the least whole number of years or steps.

Worked example 3 Paper 3

(a) Solve the equation $5^{2 x - 1} = 3^{x + 2}$ , giving your answer correct to 3 significant figures.
(b) The value of a machine is $V = 4000 \times {0.8}^{n}$ dollars, $n$ years after it is bought. Find the smallest whole number of years after which the value is less than 1000 dollars.

(a) Take logarithms: $(2 x - 1) \ln 5 = (x + 2) \ln 3$ .
Collect $x$ terms: $x (2 \ln 5 - \ln 3) = 2 \ln 3 + \ln 5$ , so $x = \frac{2 \ln 3 + \ln 5}{2 \ln 5 - \ln 3} = \frac{\ln 45}{\ln \frac{25}{3}} = 1.80$ (3 s.f.).
(b) $4000 \times {0.8}^{n} < 1000$ , so ${0.8}^{n} < 0.25$ and $n \ln 0.8 < \ln 0.25$ .
$\ln 0.8 < 0$ , so dividing by it reverses the inequality: $n > \frac{\ln 0.25}{\ln 0.8} = 6.21$ .
The smallest whole number is $n = 7$ . Check: $4000 \times {0.8}^{6} = 1048.6$ and $4000 \times {0.8}^{7} = 838.9$ .

Answer: (a) $x = 1.80$ (b) 7 years

Check yourself Paper 3

The solution of the inequality $5 \times 3^{2 x + 1} > 40$ is $x > a$ . Find $a$ , correct to 3 significant figures.

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Reducing a relationship to linear form Paper 3

Experimental data that follow a power law or an exponential law plot as a curve, but taking logarithms turns either into a straight line, whose gradient and intercept give the constants.

Power law $y = k x^{n}$ : $\ln y = \ln k + n \ln x$ . Plot $\ln y$ against $\ln x$ : a straight line with gradient $n$ and intercept $\ln k$ on the $\ln y$ axis.
Exponential law $y = k (a^{x})$ : $\ln y = \ln k + x \ln a$ . Plot $\ln y$ against $x$ : a straight line with gradient $\ln a$ and intercept $\ln k$ .

Compare with $Y = m X + c$ to see which variables go on the axes. The vertical intercept is $\ln k$ , not $k$ , so $k = e^{c}$ ; and in the exponential law the gradient is $\ln a$ , so $a = e^{m}$ . Logarithms to base 10 work in the same way, with $10^{c}$ in place of $e^{c}$ .

The gradient comes from two points on the line: read them carefully from the axes, which are $\ln x$ and $\ln y$ , not $x$ and $y$ . If the line is given by its equation, such as $\ln y = 0.8 + 1.5 \ln x$ , read off $n = 1.5$ and $\ln k = 0.8$ directly.

A straight line of $\ln y$ against $x$ means $y = k (a^{x})$ .

Worked example 4 Paper 3

The variables $x$ and $y$ satisfy $y = k (a^{x})$ , where $k$ and $a$ are constants. The graph of $\ln y$ against $x$ is a straight line passing through the points $(2, 3.10)$ and $(6, 4.30)$ , as shown in the diagram. Find the values of $k$ and $a$ , correct to 3 significant figures.

Taking logarithms: $\ln y = \ln k + x \ln a$ , a line with gradient $\ln a$ and intercept $\ln k$ .
Gradient: $\ln a = \frac{4.30 - 3.10}{6 - 2} = 0.3$ , so $a = e^{0.3} = 1.35$ (3 s.f.).
Intercept: $3.10 = \ln k + 2 (0.3)$ , so $\ln k = 2.5$ and $k = e^{2.5} = 12.2$ (3 s.f.).

Answer: $k = 12.2$ , $a = 1.35$

Check yourself Paper 3

The variables $x$ and $y$ satisfy $\ln y = 0.8 + 1.5 \ln x$ . Given that $y = k x^{n}$ , write down $n$ and find $k$ , correct to 3 significant figures.

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Key points Paper 3

$\log_{a} x = y ⟺ a^{y} = x$ ; $\log_{a} a = 1$ and $\log_{a} 1 = 0$ .
$\log (x y) = \log x + \log y$ , $\log \frac{x}{y} = \log x - \log y$ , $\log x^{k} = k \log x$ .
$\ln x$ and $e^{x}$ are inverse functions: their graphs reflect in $y = x$ . $\ln x$ is defined only for $x > 0$ .
$y = e^{k x}$ passes through $(0, 1)$ with asymptote $y = 0$ : growth for $k > 0$ , decay for $k < 0$ .
To solve for an unknown index, take logarithms; with $a^{2 x}$ and $a^{x}$ , substitute $u = a^{x} > 0$ .
Dividing an inequality by a negative logarithm, such as $\ln 0.8$ , reverses it.
$y = k x^{n}$ : $\ln y$ against $\ln x$ has gradient $n$ . $y = k (a^{x})$ : $\ln y$ against $x$ has gradient $\ln a$ . Both have intercept $\ln k$ .

Where marks are lost Paper 3

Writing $\log (x + 2)$ as $\log x + \log 2$ , or $\frac{\log 8}{\log 2}$ as $\log 4$ .
Not rejecting a solution that makes the argument of a logarithm zero or negative.
Keeping a negative root of $u = e^{x}$ or $u = 2^{x}$ : these are always positive.
Forgetting to reverse the inequality after dividing by a negative logarithm.
Rounding $\ln$ values early, so the final answer is wrong at 3 significant figures.
Giving $k$ as the intercept itself instead of $e^{intercept}$ , or $a$ as the gradient instead of $e^{gradient}$ .
Reading a point off a $\ln y$ against $\ln x$ graph as if the axes were $x$ and $y$ .

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An original lesson written to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026 to 2027 and 2028 to 2030). The worked examples and questions are our own, not Cambridge questions. Back to every AS & A Level Mathematics 9709 lesson.