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Solve $x y^m = y x^3$ for $m$.

I know this has to do with logarithms, but I'm not able to figure out how it relates to logs.

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    $\begingroup$ Welcome to Mathematics Stack Exchange. Can you take the logarithm of both sides of the equation? $\endgroup$ – J. W. Tanner Feb 2 '20 at 20:46
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    $\begingroup$ Can you divide both sides by $xy$? $\endgroup$ – Narasimham Feb 2 '20 at 20:51
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$xy^m=yx^3\implies \log x + m \log y=\log y + 3 \log x \implies (m-1)\log y=2\log x$

$\implies m-1=\dfrac{2\log x}{\log y}\implies m=\dfrac{2\log x}{\log y}+1$

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  • $\begingroup$ Thank you. Exactly what I needed to understand it. $\endgroup$ – B.Bart Feb 2 '20 at 22:34
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Rewrite $xy^m = yx^3$ as

$$y^{m-1}=x^2$$

Take log of both sides,

$$ (m-1)\ln y = 2\ln x $$

which leads to

$$m= 2\log_y x +1$$

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