The Product Rule

Idea

Two logarithms with the same base can be added by multiplying their arguments.

The reason is that logarithms are exponents, and exponents add when powers multiply. The product rule is that fact wearing different clothes.

Rule

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

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

How it is used

1. Where it comes from

Write each logarithm as an exponent and multiply the powers.

\log_a x = m \;\Rightarrow\; a^m = x,

\log_a y = n \;\Rightarrow\; a^n = y

xy = a^m \cdot a^n = a^{m+n} \;\Rightarrow\; \log_a(xy) = m + n

And $m + n$ is exactly $\log_a x + \log_a y$ .

2. Adding to get a value you can read

Separately the two terms are awkward. Together they collapse.

\log_2 4 + \log_2 8 = \log_2 32 = 5

\log 25 + \log 4 = \log 100 = 2

\log_3 6 + \log_3 4.5 = \log_3 27 = 3

Neither $\log_3 6$ nor $\log_3 4.5$ is a nice number on its own. The product is.

3. Splitting in the other direction

The rule runs both ways. Break an argument into factors when one of them is a power of the base.

\log_2 40 = \log_2(8 \cdot 5) = 3 + \log_2 5

Choose the factors so that one becomes a whole number.

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.

4. The bases must match

Expression	Combines?
$\log_5 3 + \log_5 7$	yes, into $\log_5 21$
$\log_5 3 + \log_2 7$	no, different bases
$\log 4 + \log 25$	yes, both are base $10$
$\ln 2 + \ln 6$	yes, both are base $e$

Two things the rule does not say:

\log_a x + \log_a y \ne \log_a(x + y),

\log_a x \cdot \log_a y \ne \log_a(xy)

Worked example

Evaluate $\log_6 4 + \log_6 9 + \log_2 40 - \log_2 5$ .

Remember

$\log_a x + \log_a y = \log_a(xy)$ , same base only $\log_a a^n = n$ , so write the argument as a power of the base work base by base, then add the values

The first two share base $6$ . Multiply their arguments.

\log_6 4 + \log_6 9 = \log_6 36 = 2

The last two share base $2$ . Split $40$ so that a power of $2$ appears.

\log_2 40 = \log_2(8 \cdot 5) = 3 + \log_2 5

Now the stray $\log_2 5$ cancels against the term that follows it.

3 + \log_2 5 - \log_2 5 = 3

Add the two finished values.

2 + 3 = 5

Explore

Drag the probe along the green curve and watch the height at two different values of $x$ .

Read the height at $x = 2$ , then at $x = 4$ . With the base at $2$ those heights are $1$ and $2$ . Now find $x = 8$ , the product of the two: its height is $3$ , the sum of the other two. That is the product rule seen as an addition of heights. Multiplying along the bottom adds along the side, which is the whole reason logarithms were invented.