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Math Help - Solve for variable exponent...

  1. #1
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    Solve for variable exponent...

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

    32/48 = (6^b)/(14^b)
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  2. #2
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    Quote Originally Posted by mcsquared View Post
    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
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  3. #3
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    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)
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  4. #4
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    Keep it SIMPLE:

    (6/14)^b = 32/48

    (3/7)^b = 2/3

    b = log(2/3) / log(3/7)
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  5. #5
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    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.
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