A-level Maths revision · Pure, year 1

Logarithms

The idea

log⁡ab=c\log_a b = c means ac=ba^c = b. The laws: log⁡a+log⁡b=log⁡ab\log a + \log b = \log ab, log⁡a−log⁡b=log⁡ab\log a - \log b = \log\frac{a}{b} and log⁡an=nlog⁡a\log a^n = n\log a. Take logs of both sides to solve ax=ba^x = b.

  1. Combine logs into a single log using the laws.
  2. Turn log⁡a(… )=c\log_a(\dots) = c into (… )=ac(\dots) = a^c.
  3. For ax=ba^x = b: x=ln⁡bln⁡ax = \frac{\ln b}{\ln a}. Check every log argument is positive.

Watch out log⁡(a+b)\log(a + b) does not split into log⁡a+log⁡b\log a + \log b.

See it worked

Worked example

Solve 3x=203^x = 20 to 3 significant figures.

  1. Take ln⁡\ln of both sides: xln⁡3=ln⁡20x\ln 3 = \ln 20.
  2. x=ln⁡20ln⁡3x = \frac{\ln 20}{\ln 3}.
  3. x=2.73x = 2.73 (3 s.f.).

Answer x=2.73x = 2.73

Your turn

Question 1Non-calculator3 marks

Find log⁡232\log_2 32.

Question 2Calculator allowed3 marks

Solve 5x=405^x = 40, giving xx to 3 significant figures.

Question 3Non-calculator3 marks

Solve log⁡3x+log⁡3(x−6)=3\log_3 x + \log_3(x - 6) = 3.

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