How should I go about solving this? Is it even possible?

32/48 = (6^b)/(14^b)

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- Oct 29th 2009, 07:53 PMmcsquaredSolve for variable exponent...
How should I go about solving this? Is it even possible?

32/48 = (6^b)/(14^b) - Oct 29th 2009, 07:59 PMartvandalay11

I hope you know logarithms and the rules of logs

$\displaystyle \frac{32}{48}=\frac{6^b}{14^b}$ so

$\displaystyle \log\frac{32}{48}=\log\frac{6^b}{14^b}$ so

$\displaystyle \log\frac{32}{48}=\log 6^b-\log 14^b$

$\displaystyle \log\frac{32}{48}=b\log 6-b \log 14=b(\log 6-\log 14)$

$\displaystyle \log\frac{32}{48}=b(\log\frac{6}{14})$

so yeah... gotta use logarithms - Oct 29th 2009, 08:19 PMJoelyks
32/48=6^b/14^b

32*14^b=14^b*48

32*2^b*7^b=2^b*3^b*48

32*7^b=3^b*48

32/48=(3/7)^b

lg(32/48)=lg(3/7)^b

=blg(3/7)

b=lg(32/48)/lg(3/7)

= 0.479(3s.f) - Oct 29th 2009, 08:27 PMWilmer
Keep it SIMPLE:

(6/14)^b = 32/48

(3/7)^b = 2/3

b = log(2/3) / log(3/7) - Oct 29th 2009, 10:44 PMmcsquared
Oh man, that was kinda obvious. I thought to simplify the 6/14 but the exponent being a variable made me doubt myself. D'oh!

Thank you much for the solutions, everyone. :D