The recipe: To find log_a b: ask "a to what power gives b?".
Write b as a power of a — the exponent is your answer.
Why? loga b is defined as that exponent.
IB-style question — three logs by hand
Evaluate (a) log₃ 81, (b) log₂ (1/16), (c) log₉ 3.
Step by step
- (a) Write 81 as a power of 3.
- (b) A reciprocal gives a negative power: 1/16 = 2⁻⁴.
- (c) A root gives a fractional power: 3 = 91/2.
Final answer
(a) 4 (b) −4 (c) ½.
Three quick results: log_a 1 = 0 (because a⁰ = 1).
log_a a = 1 (because a¹ = a).
log_a (1/b) = −log_a b (a reciprocal flips the sign).
IB-style question
Evaluate (a) log₇ 1, (b) log₅ 5, (c) log₂ (1/8).
Step by step
- (a) Anything to the power 0 is 1.
- (b) The base to the power 1 is itself.
- (c) 1/8 = 2⁻³.
Final answer
(a) 0 (b) 1 (c) −3.
Learn what examiners really want
See exactly what to write to score full marks. Our AI shows you model answers and the key phrases examiners look for.
Try AI Feedback Free7-day free trial • No card required
Paper 1 by hand, Paper 2 on the GDC: On Paper 1 evaluate from the definition (write the number as a power of the base).
On Paper 2 just type log or ln. For other bases on Paper 2, use the change-of-base rule (see Laws of logarithms, 1.7).
Common mistakes
- Thinking loga 1 = 1 (it's 0).
- Reading loga b as a ÷ b.
- Missing fractional or negative powers.
Do this instead
- loga 1 = 0 always (a⁰ = 1).
- loga b is the exponent that turns a into b.
- 1/16 = 2⁻⁴, so log₂(1/16) = −4.