Logarithmic and Exponential Functions

Revise Logarithmic and Exponential Functions for Pure Mathematics 2 P2 (AS Level) — revision notes and instant AI marking.

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Cambridge International AS & A Level Mathematics 9709 — Pure Mathematics 2

Logarithmic and Exponential Functions

The big idea: A logarithm is a power: log_a x = y means a^y = x. Taking logs brings an unknown down from the index, and turns curved relationships into straight lines.

Summary — What This Chapter Covers

log_a x = y ⇔ a^y = x; ln x means log_e x.
Laws: log ab = log a + log b, log(a/b) = log a − log b, log aⁿ = n log a.
e^x and ln x are inverse functions; their graphs are reflections in y = x.
Take logs of both sides to solve equations and inequalities with the unknown in the index.
y = kxⁿ becomes ln y = ln k + n ln x; y = ka^x becomes ln y = ln k + x ln a.

1. Laws of Logarithms and e^x

Logs and exponentials

Logs are indices

The laws of logs come straight from the laws of indices. They hold for any base, including e. Since e^x and ln x are inverses, e^{ln x} = x (x > 0) and ln(e^x) = x. The graph of y = e^kx passes through (0, 1); it grows for k > 0 and decays towards the x-axis for k < 0. y = ln x passes through (1, 0) and exists only for x > 0.

Laws log ab = log a + log b · log(a/b) = log a − log b · log aⁿ = n log a · log_a a = 1, log_a 1 = 0

Worked Example

Solve log₂(x + 3) + log₂(x − 3) = 4.

Practice Question 1

Solve ln(2x + 1) = 3, giving x to 3 s.f.

Practice Question 2

Solve 2 ln x − ln(x + 2) = ln 3.

2. Solving Equations and Linear Forms

Applications

Unknown indices and straight-line graphs

When the unknown is in an index, take ln (or log₁₀) of both sides and use log aⁿ = n log a. In an inequality, remember that ln of a number between 0 and 1 is negative: dividing by it reverses the inequality.

To find constants in an experimental law, take logs to get a straight line. The gradient and intercept of the line give the constants.

Linear forms y = kxⁿ ⇒ ln y = n ln x + ln k (plot ln y against ln x: gradient n, intercept ln k)
y = ka^x ⇒ ln y = x ln a + ln k (plot ln y against x: gradient ln a, intercept ln k)

Worked Example

Solve 3^x+1 = 4^2x−1, giving x to 3 s.f.

Worked Example

Find the smallest integer n for which 0.8ⁿ < 0.01.

Practice Question 1

Variables satisfy y = kxⁿ. The graph of ln y against ln x is a straight line through (1, 2.5) and (3, 6.5). Find n and k.

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Logarithmic and Exponential Functions