Logs

**Tiger** · January 13th 2010, 04:22 AM

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

**Grandad** · January 13th 2010, 04:38 AM

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)

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

Grandad

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