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So I was teaching a student high school math and I came upon this problem:

Assuming one has more than oce solution for x, for the equation

$4^{ax} = 8.b^{x}$, find all possible solution of $a, b$.

Point is I can find the answer using trial and error, how can I algebraically solve it?

Thanks.

P.S. Using logarithms isn't allowed, only exponents!

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  • $\begingroup$ is this $$4^{ax}=8\cdot b^x$$? $\endgroup$ – Dr. Sonnhard Graubner Mar 18 '17 at 10:41
  • $\begingroup$ @Dr.SonnhardGraubner Yes! $\endgroup$ – Jishan Mar 18 '17 at 10:56
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if so we have $$2ax\ln(2)=3\ln(2)+x\ln(b)$$ and then: $$x(2a\ln(2)-\ln(3))=3\ln(2))$$ thus we get $$x=\frac{3\ln(2)}{2a\ln(2)-\ln(3)}$$ for $$b>0,2a\ln(2)-\ln(3)\ne 0$$

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  • $\begingroup$ As I said this problem features in the exponent section, using logarithms isn't allowed! $\endgroup$ – Jishan Mar 18 '17 at 10:56
  • $\begingroup$ i think in this method is to find the variable $x$ impossible $\endgroup$ – Dr. Sonnhard Graubner Mar 18 '17 at 11:09
  • $\begingroup$ I thought the same, but the textbook specifically asks the question :( $\endgroup$ – Jishan Mar 18 '17 at 11:11
  • $\begingroup$ maybe there is a typo in your textbook $\endgroup$ – Dr. Sonnhard Graubner Mar 18 '17 at 11:12

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