Year 11 / Logarithms and logarithmic functions

Ask for the exponent.

logₐ(b) is the exponent needed to make aᵡ = b. For real logarithms, b must be positive, a must be positive and a cannot be 1. Logarithms turn multiplication into addition and powers into multipliers.

A logarithmic graph is the inverse of its exponential graph. The rule logₐ(x − h) + k has vertical asymptote x = h and domain x > h. Algebraic rearrangement can produce candidates outside this domain, so check every proposed solution in the original equation.

INTERACTIVE MODEL

Try it. Watch it change.

y = log₂(x − 0) + 0

Horizontal shift h0 Vertical shift k0Input x1

f(1) = 0 · domain x > 0

The graph is calculated from the displayed rule. Drag the highlighted point or use the sliders.

Explore: Shift the logarithmic curve right. Observe which inputs disappear from its real domain.

Product law

\log_a(xy)=\log_a x+\log_a y

Change of base

\log_a x=\frac{\ln x}{\ln a}

WORKED EXAMPLE

Solve log₂(x − 1) = 3.

The domain requires x > 1.
x − 1 = $2^{3}$ = 8, so x = 9.
9 satisfies the domain and the original equation.

Assumed knowledge

Exponents and inequalities.

Learning checkpoints & source

YOUR LEARNING CHECKPOINT

Apply logarithmic laws with domain restrictions.
Solve logarithmic and exponential equations.

QCAA Mathematical Methods 2025 v1.3 · p. 21 ↗

READY TO TRY IT?

Ask for the exponent.

Solve log₂(x − 1) = 3.

Put the idea to work.