1. ## Logs

If logn(108)=a, and logn(72)=b, find logn(2)in terms of a and b.
Consider logn(32).

2. Hello Tiger
Originally Posted by Tiger
If logn(108)=a, and logn(72)=b, find logn(2)in terms of a and b.
Consider logn(32).
$\log_n(108)=a$
$\Rightarrow n^a=108=2^2\times3^3$ ...(1)
$\log_n(72) = b$
$\Rightarrow n^b=72=2^3\times3^2$ ...(2)
$x=\log_n(2)$
$\Rightarrow n^x= 2$ ...(3)
So we need to eliminate the $3$'s from equations (1) and (2), to be left with a power of $2$ only. So square both sides of (1) and cube both sides of (2):
$n^{2a} = 2^4\times 3^6$
and
$n^{3b} = 2^9\times 3^6$
Divide:
$\frac{n^{3b}}{n^{2a}}= \frac{2^9\times 3^6}{2^4\times 3^6}$

$\Rightarrow n^{3b-2a} = 2^5=\Big(n^x\Big)^5$, from
(3)

$\Rightarrow x = \tfrac15(3b-2a)$