The recipe: To find loga 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.
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Three quick results: loga 1 = 0 (because a⁰ = 1).
loga a = 1 (because a¹ = a).
loga (1/b) = −loga 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.
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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.