# Solve for variable exponent...

• October 29th 2009, 07:53 PM
mcsquared
Solve for variable exponent...
How should I go about solving this? Is it even possible?

32/48 = (6^b)/(14^b)
• October 29th 2009, 07:59 PM
artvandalay11
Quote:

Originally Posted by mcsquared
How should I go about solving this? Is it even possible?

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

I hope you know logarithms and the rules of logs

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

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

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

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

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

so yeah... gotta use logarithms
• October 29th 2009, 08:19 PM
Joelyks
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)
• October 29th 2009, 08:27 PM
Wilmer
Keep it SIMPLE:

(6/14)^b = 32/48

(3/7)^b = 2/3

b = log(2/3) / log(3/7)
• October 29th 2009, 10:44 PM
mcsquared
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