# Logarithms. Solve for m

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.

• Welcome to Mathematics Stack Exchange. Can you take the logarithm of both sides of the equation? – J. W. Tanner Feb 2 '20 at 20:46
• Can you divide both sides by $xy$? – Narasimham Feb 2 '20 at 20:51

$$xy^m=yx^3\implies \log x + m \log y=\log y + 3 \log x \implies (m-1)\log y=2\log x$$

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

• Thank you. Exactly what I needed to understand it. – B.Bart Feb 2 '20 at 22:34

Rewrite $$xy^m = yx^3$$ as

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

Take log of both sides,

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

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