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Topic 6

Logarithmic and exponential functions

Syllabus 6.1–6.3 · Papers 1 and 2

🎯What you need to be able to do

  • Move between index form and logarithm form, in any base, including \( \lg \) (base 10) and \( \ln \) (base \(e\)).
  • Use the laws of logarithms, including change of base, to simplify and to combine logarithms.
  • Solve equations of the form \( a^{x} = b \), and equations involving logarithms.
  • Sketch \( y = ke^{nx} + a \) and \( y = k\ln(ax + b) \), with intercepts and the equations of asymptotes.
  • Know that \( e^{x} \) and \( \ln x \) are inverse functions.

📚The mathematics

What a logarithm is

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

A logarithm is a power: \( \log_{2} 32 = 5 \) because \( 2^{5} = 32 \). Two special ones have their own names: \( \lg x = \log_{10} x \) and \( \ln x = \log_{e} x \). You can only take the log of a positive number, which is why answers to log equations must be checked.

The laws

\( \log_{a} xy = \log_{a} x + \log_{a} y \)
\( \log_{a} \dfrac{x}{y} = \log_{a} x - \log_{a} y \)
\( \log_{a} x^{k} = k \log_{a} x \)
\( \log_{a} a = 1 \), \( \log_{a} 1 = 0 \)
\( \log_{a} b = \dfrac{\log_{c} b}{\log_{c} a} \) (change of base)
\( \log_{a} b = \dfrac{1}{\log_{b} a} \)

None of these are in the List of formulas — learn them. And note what is not a law: \( \log(x + y) \) does not split, and \( \dfrac{\log x}{\log y} \) is not \( \log x - \log y \).

To write something like \( 3 + 2\lg p - \lg q \) as a single logarithm, turn the number into a log of the same base first: \( 3 = \lg 1000 \). Then \( 3 + 2\lg p - \lg q = \lg \dfrac{1000p^{2}}{q} \).

Solving equations

  • \( a^{x} = b \): take logs of both sides (\( \ln \) or \( \lg \)), bring the power down, solve the linear equation.
  • Logs on both sides: combine each side into a single log of the same base, then drop the logs (or rewrite in index form).
  • Different bases: change them to one base first. \( \log_{4} X = \tfrac{1}{2}\log_{2} X \), because \( \log_{2} 4 = 2 \).

Then check every answer in the original: any value that makes the argument of a log zero or negative must be rejected.

Graphs and asymptotes

\( y = e^{x} \) and \( y = \ln x \) are inverses, so each is the reflection of the other in \( y = x \): \( e^{x} \) passes through \( (0, 1) \) with asymptote \( y = 0 \); \( \ln x \) passes through \( (1, 0) \) with asymptote \( x = 0 \).

The curves y = e to the x, through (0, 1) with horizontal asymptote y = 0, and y = ln x, through (1, 0) with vertical asymptote x = 0, drawn as reflections of each other in the dashed line y = x.
\( e^{x} \) and \( \ln x \) are mirror images in \( y = x \); so are their intercepts and asymptotes.

The syllabus limits the graphs to two families:

  • \( y = ke^{nx} + a \): horizontal asymptote \( y = a \) (the exponential part tends to 0 at one end); \(y\)-intercept \( k + a \).
  • \( y = k\ln(ax + b) \): vertical asymptote where \( ax + b = 0 \); crosses the \(x\)-axis where \( ax + b = 1 \).
Two panels. Left: y = 3e to the 2x minus 4, crossing the y-axis at minus 1 and the x-axis at a half ln four thirds, approaching the dashed asymptote y = minus 4 to the left. Right: y = 2 ln(x + 3), with dashed vertical asymptote x = minus 3, crossing the x-axis at minus 2 and the y-axis at 2 ln 3.
State the asymptote as an equation (\( y = -4 \), \( x = -3 \)) and label both intercepts, exactly or to 3 s.f.

✏️Worked example

(a) Solve \( 3^{2x + 1} = 5^{x} \), giving your answer to 3 significant figures. [3] (b) Solve \( \log_{2} x + \log_{2}(x - 2) = 3 \). [3] (c) Solve \( \log_{3} x - 4\log_{x} 3 = 3 \). [4]

(a) Take natural logs: \( (2x + 1)\ln 3 = x\ln 5 \). Collect the \(x\) terms: \( x(2\ln 3 - \ln 5) = -\ln 3 \), so

\[ x = \frac{-\ln 3}{2\ln 3 - \ln 5} = \frac{-\ln 3}{\ln 1.8} = -1.87 \]

(b) Combine: \( \log_{2} x(x - 2) = 3 \), so \( x(x - 2) = 2^{3} = 8 \) and \( x^{2} - 2x - 8 = 0 \), giving \( (x - 4)(x + 2) = 0 \). But \( x = -2 \) would need \( \log_{2}(-2) \), which does not exist, so \( x = 4 \) only.

(c) Change base: \( \log_{x} 3 = \dfrac{1}{\log_{3} x} \). Let \( u = \log_{3} x \):

\[ u - \frac{4}{u} = 3 \;\Longrightarrow\; u^{2} - 3u - 4 = 0 \;\Longrightarrow\; (u - 4)(u + 1) = 0 \]

So \( \log_{3} x = 4 \Rightarrow x = 81 \), or \( \log_{3} x = -1 \Rightarrow x = \tfrac{1}{3} \).

Check it. (a) With the unrounded \( x = -1.869\ldots \), both sides equal 0.0494 ✓. (b) \( \log_{2} 4 + \log_{2} 2 = 2 + 1 = 3 \) ✓. (c) At \( x = 81 \): \( 4 - 4\left(\tfrac{1}{4}\right) = 3 \) ✓; at \( x = \tfrac{1}{3} \): \( -1 - 4(-1) = 3 \) ✓.
Splitting what does not split. \( \log_{2}(x - 2) \) is not \( \log_{2} x - \log_{2} 2 \). The laws turn products and quotients into sums and differences — never the other way round for a sum inside the log.

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) Write \( 3 + 2\lg p - \lg q \) as a single logarithm. [3]
\( \lg 1000 + \lg p^{2} - \lg q = \lg \dfrac{1000p^{2}}{q} \).
2. (No calculator.) Write \( \dfrac{2}{\log_{3} e} \) as a single natural logarithm. [2]
\( \log_{3} e = \dfrac{1}{\log_{e} 3} = \dfrac{1}{\ln 3} \), so \( \dfrac{2}{\log_{3} e} = 2\ln 3 = \ln 9 \).
3. (No calculator.) Solve \( \log_{4}(x + 6) - \log_{4} x = \tfrac{1}{2} \). [3]
\( \log_{4}\dfrac{x + 6}{x} = \tfrac{1}{2} \Rightarrow \dfrac{x + 6}{x} = 4^{\frac{1}{2}} = 2 \Rightarrow x + 6 = 2x \Rightarrow x = 6 \).
4. (Calculator.) Solve \( 5^{x - 1} = 2^{x + 2} \). [3]
\( (x - 1)\ln 5 = (x + 2)\ln 2 \Rightarrow x(\ln 5 - \ln 2) = \ln 5 + 2\ln 2 \Rightarrow x = \dfrac{\ln 20}{\ln 2.5} = 3.27 \).
5. (No calculator.) Solve \( \log_{2} x = \log_{4}(x + 12) \). [4]
\( \log_{4}(x + 12) = \dfrac{\log_{2}(x + 12)}{\log_{2} 4} = \tfrac{1}{2}\log_{2}(x + 12) \). So \( 2\log_{2} x = \log_{2}(x + 12) \), \( x^{2} = x + 12 \), \( (x - 4)(x + 3) = 0 \). Reject \( x = -3 \) (\( \log_{2}(-3) \) is undefined): \( x = 4 \).
6. (No calculator.) The curve \( y = 5e^{-2x} + 3 \). (a) State the equation of the asymptote and the \(y\)-intercept. (b) Find the exact value of \(x\) for which \( y = 13 \). [4]
(a) As \( x \to \infty \), \( e^{-2x} \to 0 \), so the asymptote is \( y = 3 \); \(y\)-intercept \( (0, 8) \). (b) \( 5e^{-2x} = 10 \Rightarrow e^{-2x} = 2 \Rightarrow -2x = \ln 2 \Rightarrow x = -\tfrac{1}{2}\ln 2 \).
7. (Calculator.) The number of bacteria in a sample is modelled by \( N = 250e^{kt} \), where \(t\) is in hours. After 5 hours there are 400 bacteria. (a) Find \(k\). (b) Find the time at which there are 1000 bacteria. [5]
(a) \( 400 = 250e^{5k} \Rightarrow e^{5k} = 1.6 \Rightarrow k = \tfrac{1}{5}\ln 1.6 = 0.0940 \). (b) \( 1000 = 250e^{kt} \Rightarrow e^{kt} = 4 \Rightarrow t = \dfrac{\ln 4}{k} = \dfrac{5\ln 4}{\ln 1.6} = 14.7 \) hours. Keep \(k\) unrounded for part (b): carrying a rounded value through several steps is how a final answer drifts out of 3 s.f. accuracy.

🔗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 \( y = ke^{nx} + a \) with sliders to see how each constant moves the asymptote and intercept
  • Khan Academy — properties of logarithms and change of base