Changing the Base

Idea

Every rule so far has needed the bases to match. This one removes that obstacle: any logarithm can be rewritten in any other valid base.

That makes two things possible. A calculator with only $\log$ and $\ln$ keys can find a logarithm in any base, and two logarithms in different bases can finally be compared or combined.

Rule

$\log_a x = \dfrac{\log_b x}{\log_b a}$

$\log_a b = \dfrac{1}{\log_b a}$

How it is used

1. Where it comes from

Let $y = \log_a x$ , which means $a^y = x$ . Take $\log_b$ of both sides and pull the exponent out.

\log_b(a^y) = \log_b x \;\Rightarrow\; y \log_b a = \log_b x

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

The new base $b$ is yours to choose. Any allowed base gives the same answer.

2. Choosing the new base

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}

The second is exact because $8$ and $32$ are both powers of $2$ . Picking the right new base is the whole skill.

Exponential and logarithm

Try this. Compare bases 2 and 0.5: growth becomes decay. Move the input to 1; the logarithm is zero for either base.

BasePositive input x0.5

Base 2 · Blue: y = 2ˣ · Green: y = log_b(x) · Dashed: y = x. At x = 0.5, y = -1.

3. The upside-down case

Put $x = b$ in the rule. Since $\log_b b = 1$ :

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

So $\log_a b$ and $\log_b a$ are reciprocals, and their product is $1$ .

\log_5 3 \cdot \log_3 5 = 1

4. Chains collapse

Write each factor over a common base and watch the middles cancel.

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

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

Worked example

Evaluate $\log_2 5 \cdot \log_5 8 + \log_9 27$ .

Remember

$\log_a x = \dfrac{\log_b x}{\log_b a}$ , with $b$ any base you like $\log_a a^n = n$ $\log_a b \cdot \log_b a = 1$

Rewrite the product over base $e$ . Any common base works.

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

The $\ln 5$ cancels, leaving a single logarithm.

\frac{\ln 8}{\ln 2} = \log_2 8 = 3

For the last term choose base $3$ , since $9$ and $27$ are both powers of $3$ .

\log_9 27 = \frac{\log_3 27}{\log_3 9} = \frac{3}{2}

Add the two values.

3 + \frac{3}{2} = \frac{9}{2}

Explore

Drag the base slider and watch the same curve stretch.

Pick a value of $x$ and read its height. Now move the base from $2$ to $4$ and read the height above the same $x$ : it has halved. Try a different $x$ and it halves too. The rule says why the factor is the same everywhere:

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}

That denominator is a number, not something that depends on $x$ .

Changing the base never bends the curve into a new shape. It only scales every height by one fixed factor. Which factor depends on the two bases, but $(1, 0)$ never moves whichever you pick, because multiplying $0$ by anything leaves $0$ .