Year 12 Exponentials and logarithms: Evaluating logarithms

Evaluating a logarithm means finding the exponent that makes the base produce the given number. Exact answers come from rewriting the argument as a power of the base.

\log_{b} (a) = x

means exactly the same thing as

b^{x} = a

Main idea

If you can express the argument as $b^{k}$ , then $\log_{b} (b^{k}) = k$ .

Worked example

Problem: Evaluate

\log_{2} (32) - \log_{3} (1 / 27)

Rewrite $32$ as a power of 2: $32 = 2^{5}$ .
So $\log_{2} (32) = 5$ .
Rewrite $1 / 27$ as a power of 3: $1 / 27 = 3^{- 3}$ .
So $\log_{3} (1 / 27) = - 3$ .
Substitute into the expression: $5 - (- 3) = 8$ .

Answer:

\log_{2} (32) - \log_{3} (1 / 27) = 8

How to recognise exact evaluations

Exact logarithm questions usually hide a friendly power. The argument may already look like a power of the base, or it may need a quick rewrite. Fractions often become negative powers, square roots become power one half, and reciprocal roots become negative fractional powers. For example, $\log_{5} (1 / 125)$ is easier if you first write $1 / 125 = 5^{- 3}$ . The logarithm then simply reads off the exponent.

Worked example 2

Problem: Evaluate

\log_{4} (1 / 2) + \log_{9} (3)

Rewrite $1 / 2$ in terms of base 4. Since $4 = 2^{2}$ , we can write $1 / 2 = 4^{- 1 / 2}$ .
So $\log_{4} (1 / 2) = - 1 / 2$ .
Now rewrite $3$ in terms of base 9. Because $9 = 3^{2}$ , we have $3 = 9^{1 / 2}$ .
So $\log_{9} (3) = 1 / 2$ .
Add the values: $- 1 / 2 + 1 / 2 = 0$ .

Answer: the value is

0

Common mistakes

Rewriting the base instead of the argument and then stopping halfway.
Forgetting that reciprocal values give negative exponents.
Mixing up $\log_{b} (a)$ with $\log_{a} (b)$ .
Converting an exact logarithm into a decimal unnecessarily.

Quick check

After you get an answer, convert it back into exponential form and verify it mentally. If you say $\log_{4} (1 / 2) = - 1 / 2$ , then you should also be able to confirm that $4^{- 1 / 2} = 1 / 2$ . That reverse check is fast and catches sign errors immediately.

Revision focus

Exponential and logarithmic questions become much easier when you classify the task before starting. Ask whether you are evaluating a logarithm exactly, simplifying with log laws, solving an exponential equation, solving a logarithmic equation, or modelling growth and decay. Each of those tasks has a standard pattern, and confusion usually comes from mixing those patterns together rather than from the algebra itself.

Another high-value habit is to reverse-check the answer. If a logarithm equals a certain number, convert it back into exponential form and see whether the statement is true. If a model gives a time or population value, ask whether the rounding and units make sense in context. Domain restrictions, positive log arguments, and sensible growth factors are not side details in this topic. They are part of what makes the answer mathematically valid.

A quick final check for this topic is to ask whether the form of the answer matches the form of the question. Exact evaluations should usually stay exact. Solving questions should respect domain restrictions. Modelling questions should include sensible units, rounding, and interpretation. Those checks are short, but they are exactly the checks that stop a nearly correct solution from becoming an incorrect one.

Practice links

Evaluation of logarithms