P2/3 2 Logarithmic and exponential functions | A Level Maths 9709

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🎯What you need to be able to do

Use the laws of logarithms fluently in both directions.
Understand \( \mathrm{e}^{x} \) and \( \ln x \) as inverse functions, and sketch both graphs.
Solve equations of the form \( a^{x} = b \) by taking logarithms.
Solve equations and inequalities involving \( \mathrm{e}^{x} \) and \( \ln x \), including hidden quadratics.
Reduce \( y = ax^{n} \) and \( y = ab^{x} \) to linear form and find the constants from a graph.
Know the domain restrictions and check solutions against them.

📚The mathematics

What a logarithm is

A logarithm is an exponent, and that one sentence is the whole idea:

\[ a^{x} = b \iff \log_a b = x \qquad (a > 0,\ a \ne 1,\ b > 0) \]

So \( \log_2 8 = 3 \) simply records that \( 2^{3} = 8 \). The natural logarithm \( \ln x \) is \( \log_{\mathrm{e}} x \), where \( \mathrm{e} \approx 2.718 \).

The laws

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

\( \log_a x - \log_a y = \log_a \dfrac{x}{y} \)

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

\( \log_a a = 1 \), \( \log_a 1 = 0 \)

You need these in both directions. Forwards, \( \ln 12 = \ln 4 + \ln 3 \). Backwards, \( 2\ln x - \ln y \) collapses to \( \ln\!\left(\dfrac{x^{2}}{y}\right) \). Collapsing a sum of logarithms into a single logarithm is how you solve almost every logarithmic equation, because once you have \( \log_a A = \log_a B \) you may conclude \( A = B \).

Three laws that do not exist. \( \log(x+y) \) is not \( \log x + \log y \). \( \dfrac{\log x}{\log y} \) is not \( \log\dfrac{x}{y} \). And \( (\log x)^{2} \) is not \( 2\log x \). Each of these appears in examiner reports every session.

The graphs, and the inverse relationship

\( y = \mathrm{e}^{x} \) passes through \( (0,1) \), rises steeply, and has the \(x\)-axis as a horizontal asymptote — it is never negative and never zero. \( y = \ln x \) passes through \( (1,0) \), rises ever more slowly, and has the \(y\)-axis as a vertical asymptote; it is defined only for \( x > 0 \).

Each is the other reflected in \( y = x \), which is a fast way to recall which asymptote goes where. The relationships worth having automatic:

\( \ln\left(\mathrm{e}^{x}\right) = x \)

\( \mathrm{e}^{\ln x} = x \)

\( a^{x} = \mathrm{e}^{x\ln a} \)

Solving equations

Exponential equations. Take logarithms of both sides and use the power law to bring the unknown down from the exponent:

\[ 3^{x} = 20 \;\Longrightarrow\; x\ln 3 = \ln 20 \;\Longrightarrow\; x = \frac{\ln 20}{\ln 3} = 2.73 \]

When the unknown appears in both exponents, as in \( 5^{2x-1} = 3^{x+2} \), take logarithms immediately and then expand both sides before collecting the \(x\) terms.

Hidden quadratics. \( \mathrm{e}^{2x} - 5\mathrm{e}^{x} + 6 = 0 \) is a quadratic in \( \mathrm{e}^{x} \), since \( \mathrm{e}^{2x} = \left(\mathrm{e}^{x}\right)^{2} \). Substitute \( u = \mathrm{e}^{x} \), solve, then undo. Crucially, reject any non-positive root for \(u\): \( \mathrm{e}^{x} \) is always positive, so \( \mathrm{e}^{x} = -2 \) has no solution. Rejecting it with that reason is a mark.

Logarithmic equations. Collapse to a single logarithm, remove the logarithm, then check every answer against the domain. Solving \( \ln(x) + \ln(x-3) = \ln 4 \) gives \( x^{2} - 3x - 4 = 0 \), so \( x = 4 \) or \( x = -1 \) — but \( x = -1 \) would require \( \ln(-1) \), so it must be discarded. The algebra cannot tell you that; only the domain check can.

Reducing to linear form

This is the sub-topic that produces whole examination questions, and the method is mechanical once you see it. Take logarithms of both sides and compare with \( Y = mX + c \).

Power law, \( y = ax^{n} \):

\[ \ln y = \ln a + n\ln x \]

Plotting \( \ln y \) against \( \ln x \) gives a straight line of gradient \(n\) and intercept \( \ln a \).

Exponential law, \( y = ab^{x} \):

\[ \ln y = \ln a + x\ln b \]

Plotting \( \ln y \) against \( x \) gives a straight line of gradient \( \ln b \) and intercept \( \ln a \).

Note the difference: the power law needs logarithms on both axes, the exponential law on the \(y\)-axis only. Which plot is straight therefore identifies which model applies.

The intercept is \( \ln a \), not \(a\). Having read the intercept off the graph you must exponentiate to recover the constant: \( a = \mathrm{e}^{\text{intercept}} \). The same applies to the gradient in the exponential case, where \( b = \mathrm{e}^{\text{gradient}} \). Omitting that final step is the single most common error on reduction-to-linear-form questions.

✏️Worked example

(a) Solve \( \mathrm{e}^{2x} - 7\mathrm{e}^{x} + 12 = 0 \), giving answers in exact form. (b) Solve \( \ln(2x + 1) = 3 \), giving your answer to 3 significant figures. (c) The variables \(x\) and \(y\) satisfy \( y = ab^{x} \). When \( \ln y \) is plotted against \(x\), a straight line passing through \( (0, 1.6) \) and \( (5, 4.1) \) is obtained. Find \(a\) and \(b\).

(a) Let \( u = \mathrm{e}^{x} \). Then \( u^{2} - 7u + 12 = 0 \), so \( (u-3)(u-4) = 0 \) and \( u = 3 \) or \( u = 4 \). Both are positive, so both are valid:

\[ \mathrm{e}^{x} = 3 \Rightarrow x = \ln 3, \qquad \mathrm{e}^{x} = 4 \Rightarrow x = \ln 4 \]

The question says exact, so leave the answers as \( \ln 3 \) and \( \ln 4 \).

(b) Exponentiate both sides: \( 2x + 1 = \mathrm{e}^{3} = 20.0855 \), so \( 2x = 19.0855 \) and \( x = 9.54 \) to 3 s.f. Check the domain: \( 2x + 1 = 20.09 > 0 \), so the logarithm is defined and the answer stands.

(c) Taking logarithms, \( \ln y = \ln a + x\ln b \), which is a straight line with \( \ln y \) on the vertical axis and \(x\) on the horizontal.

The gradient is \( \ln b \):

\[ \ln b = \frac{4.1 - 1.6}{5 - 0} = \frac{2.5}{5} = 0.5 \;\Longrightarrow\; b = \mathrm{e}^{0.5} = 1.65 \]

The intercept is \( \ln a \), and the line passes through \( (0, 1.6) \), so \( \ln a = 1.6 \) and \( a = \mathrm{e}^{1.6} = 4.95 \).

Check it. With \( a = 4.95 \) and \( b = 1.65 \), test the second data point: \( y = 4.95 \times 1.65^{5} = 4.95 \times 12.18 = 60.3 \), and \( \ln 60.3 = 4.10 \). ✓ That matches the given point \( (5, 4.1) \), confirming both constants. In part (a), a quick check: \( \ln 3 = 1.10 \) and \( \mathrm{e}^{2.20} - 7\mathrm{e}^{1.10} + 12 = 9 - 21 + 12 = 0 \). ✓

Exact means exact, and the intercept needs undoing. In (a), writing \( x = 1.10 \) and \( x = 1.39 \) when the question said exact form loses the marks despite correct method. In (c), stopping at \( \ln a = 1.6 \) and quoting \( a = 1.6 \) is the classic error — the graph gives you the logarithm of the constant, never the constant itself.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Solve \( 5^{2x-1} = 12 \), giving your answer to 3 significant figures.

Take natural logarithms: \( (2x-1)\ln 5 = \ln 12 \), so \( 2x - 1 = \dfrac{\ln 12}{\ln 5} = \dfrac{2.4849}{1.6094} = 1.5440 \). Then \( 2x = 2.5440 \) and \( x = 1.27 \). Note the whole exponent \( 2x-1 \) comes down as a bracket — dropping the bracket and writing \( 2x\ln 5 - 1 \) is the usual slip.

2. Solve \( \log_3(x + 6) - \log_3(x) = 2 \).

Combine using the subtraction law: \( \log_3\!\left(\dfrac{x+6}{x}\right) = 2 \), so \( \dfrac{x+6}{x} = 3^{2} = 9 \). Then \( x + 6 = 9x \), giving \( 8x = 6 \) and \( x = 0.75 \). Domain check: both \( x = 0.75 > 0 \) and \( x + 6 = 6.75 > 0 \), so the solution is valid.

3. Solve \( \mathrm{e}^{2x} - \mathrm{e}^{x} - 6 = 0 \).

Let \( u = \mathrm{e}^{x} \): \( u^{2} - u - 6 = 0 \), so \( (u-3)(u+2) = 0 \) and \( u = 3 \) or \( u = -2 \). Reject \( u = -2 \), since \( \mathrm{e}^{x} > 0 \) for all real \(x\) — stating that reason earns the mark. So \( \mathrm{e}^{x} = 3 \) and \( x = \ln 3 = 1.10 \).

4. Solve the inequality \( \ln(3x - 2) < 2 \).

Exponentiating (a strictly increasing operation, so the inequality direction is unchanged): \( 3x - 2 < \mathrm{e}^{2} = 7.389 \), so \( 3x < 9.389 \) and \( x < 3.13 \). But the logarithm also requires \( 3x - 2 > 0 \), that is \( x > \tfrac{2}{3} \). Combining, \( \tfrac{2}{3} < x < 3.13 \). The lower bound comes only from the domain and is missed by candidates who treat this as pure algebra.

5. The variables satisfy \( y = ax^{n} \). A plot of \( \lg y \) against \( \lg x \) is a straight line through \( (0, 0.4) \) with gradient \( -1.5 \). Find \(a\) and \(n\).

Taking logarithms base 10: \( \lg y = \lg a + n\lg x \). The gradient is \(n\), so \( n = -1.5 \). The intercept is \( \lg a = 0.4 \), so \( a = 10^{0.4} = 2.51 \). Note this uses base 10 rather than natural logarithms — the structure is identical, but the un-logging step uses \( 10^{\,} \) rather than \( \mathrm{e}^{\,} \).

6. Given \( \ln y = 2\ln x - \ln(x + 1) + \ln 3 \), express \(y\) in terms of \(x\).

Collapse the right-hand side into a single logarithm. The power law turns \( 2\ln x \) into \( \ln x^{2} \), then addition multiplies and subtraction divides: \( \ln y = \ln\!\left(\dfrac{3x^{2}}{x+1}\right) \). Since \( \ln \) is one-one, \( y = \dfrac{3x^{2}}{x+1} \). Working the laws backwards like this is exactly as examinable as working them forwards.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Desmos — plot \( \mathrm{e}^{x} \) and \( \ln x \) with \( y = x \) dashed to see the reflection
Khan Academy — logarithm laws and exponential equations
Better Explained — an intuitive account of \( \mathrm{e} \) and the natural logarithm