]> Exercise 2.13

## Exercise 2.13

Deduce them, that is, deduce: $ln ⁡ a b = ln ⁡ a + ln ⁡ b$ , and $( log ⁡ a b ) * ( log ⁡ b c ) = log ⁡ a c$ .

Solution:

The natural logarithm function $ln ⁡ x$ is inverse to the exponential function. If in the first equation we take the exponential function of both sides we get $exp ⁡ ( ln ⁡ a b )$ on the left, which is $a b$ , and $exp ⁡ ( ln ⁡ a + ln ⁡ b )$ on the right.

Using the identity $exp ⁡ ( s + t ) = ( exp ⁡ s ) * ( exp ⁡ t )$ the right hand side becomes $a b$ as well.

The identity $exp ⁡ s t = ( exp ⁡ s ) t$ implies the second identity in the case $a = exp ⁡ 1$ by the following argument:
in that case $log ⁡ a b$ is ln b and $log ⁡ a c = ln ⁡ c$ .
We can then apply the exponential function to both sides of the equation we want to prove and we get

$exp ⁡ ( ( ln ⁡ b ) * ( log ⁡ b c ) ) = exp ⁡ ( ln ⁡ c ) = c$

By the identity the left hand side here becomes $exp ⁡ ( ln ⁡ b ) ^ log ⁡ b c$ , or $b ^ log ⁡ b c$ which is again $c$ .

But this tells us that in general

$log ⁡ b c = ln ⁡ c ln ⁡ b$

and the general claim, $( log ⁡ a b ) * ( log ⁡ b c ) = log ⁡ a c$ follows immediately from this fact, by substitution.