Home  18.013A  Chapter 2  Section 2.4 


Deduce them, that is, deduce: $\mathrm{ln}ab=\mathrm{ln}a+\mathrm{ln}b$ , and $({\mathrm{log}}_{a}b)*({\mathrm{log}}_{b}c)={\mathrm{log}}_{a}c$ .
Solution:
The natural logarithm function $\mathrm{ln}x$ is inverse to the exponential function. If in the first equation we take the exponential function of both sides we get $\mathrm{exp}(\mathrm{ln}ab)$ on the left, which is $ab$ , and $\mathrm{exp}(\mathrm{ln}a+\mathrm{ln}b)$ on the right.
Using the identity $\mathrm{exp}(s+t)=(\mathrm{exp}s)*(\mathrm{exp}t)$ the right hand side becomes $ab$ as well.
The identity
$\mathrm{exp}st={(\mathrm{exp}s)}^{t}$
implies the second identity in the case
$a=\mathrm{exp}1$
by the following argument:
in that case
${\mathrm{log}}_{a}b$
is ln b and
${\mathrm{log}}_{a}c=\mathrm{ln}c$
.
We can then apply the exponential function to both sides of the equation we want to prove and we get
By the identity the left hand side here becomes $\mathrm{exp}(\mathrm{ln}b)^{\mathrm{log}}_{b}c$ , or $b^{\mathrm{log}}_{b}c$ which is again $c$ .
But this tells us that in general
and the general claim, $({\mathrm{log}}_{a}b)*({\mathrm{log}}_{b}c)={\mathrm{log}}_{a}c$ follows immediately from this fact, by substitution.
