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Please let me know how to solve this equation:

$$100^{1-b}=\frac{1}{2}40^{1-b}+\frac{1}{2}200^{1-b}$$

I try to use the trick of $x=e^{\log x}$ But it doesn't work

And $b\not=1$

Kenneth Chen
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1 Answers1

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$$100^{1-b}=\frac 12 40^{1-b}+\frac 12 200^{1-b}\\ 100^{1-b}=\frac 1240^{1-b}+\frac 12\cdot 2^{1-b}\cdot100^{1-b}\\ 100^{1-b}(1-2^{-b})=\frac 1240^{1-b}\\5^{1-b}(1-2^{-b})=2^{-b}\\ 2^b-1=5^{b-1}$$

Alpha finds two solutions, $b=1$ and $b\approx 0.650919$ The first is easy to verify, the second I don't think you will find without numerics.

Ross Millikan
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  • Thanks Ross, this is kind of weird as I agree that the answer should be found with numerical method, however this question was asked in a rather theoretical-based course, which made me confused – Kenneth Chen Oct 01 '13 at 01:34