Exercise 2.13

]> Exercise 2.13

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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 = \frac{\ln c}{\ln b}

and the general claim, $(\log_{a} b) * (\log_{b} c) = \log_{a} c$ follows immediately from this fact, by substitution.