Changing the Base — Cheat sheet

The rule

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

Argument on top, old base underneath. The new base $b$ is yours to choose.

$y = \log_a x$ means $a^y = x$ . Take $\log_b$ of both sides:

y \log_b a = \log_b x \;\Rightarrow\; y = \frac{\log_b x}{\log_b a}

Goal	Choose
use a calculator	$b = 10$ or $b = e$
get an exact value	a base both numbers are powers of

\log_2 7 = \frac{\ln 7}{\ln 2},

\log_8 32 = \frac{\log_2 32}{\log_2 8} = \frac{5}{3},

\log_9 27 = \frac{3}{2}

Put $x = b$ and use $\log_b b = 1$ :

\log_a b = \frac{1}{\log_b a}

\Longrightarrow

\log_a b \cdot \log_b a = 1

So $\log_5 3 \cdot \log_3 5 = 1$ , and if $\log_a b = 4$ then $\log_b a = \dfrac{1}{4}$ . Reciprocal, never negative.

Put every factor over one base and the middles cancel.

\log_2 5 \cdot \log_5 8 = \frac{\ln 5}{\ln 2} \cdot \frac{\ln 8}{\ln 5} = \log_2 8 = 3

\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdot \log_5 8 = \log_2 8 = 3

A chain keeps only the first base and the last argument.

$\dfrac{\ln 3}{\ln 7}$ is $\log_7 3$ , not $\log_3 7$ . Check which one is underneath.
$\dfrac{\log_a x}{\log_a y}$ is not $\log_a \dfrac{x}{y}$ . That is the quotient rule, a different thing.