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).
$\displaystyle \log_n(108)=a$$\displaystyle \Rightarrow n^a=108=2^2\times3^3$ ...(1)
$\displaystyle \log_n(72) = b$$\displaystyle \Rightarrow n^b=72=2^3\times3^2$ ...(2)
$\displaystyle x=\log_n(2)$$\displaystyle \Rightarrow n^x= 2$ ...(3)
So we need to eliminate the $\displaystyle 3$'s from equations (1) and (2), to be left with a power of $\displaystyle 2$ only. So square both sides of (1) and cube both sides of (2):$\displaystyle n^{2a} = 2^4\times 3^6$
and$\displaystyle n^{3b} = 2^9\times 3^6$
Divide:$\displaystyle \frac{n^{3b}}{n^{2a}}= \frac{2^9\times 3^6}{2^4\times 3^6}$
$\displaystyle \Rightarrow n^{3b-2a} = 2^5=\Big(n^x\Big)^5$, from (3)
$\displaystyle \Rightarrow x = \tfrac15(3b-2a)$
Grandad