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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.

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