# Finding the inverse of a log problem

I have a homework problem that I am struggling to understand. The problem is Find a formula for the inverse function $f^ {-1}$ of the function $f$. $$f(x)=\log_{2x}3$$

Here is my attempt at solving this problem. $$y=\log_{2x}3$$ $$2x ^y=3$$ $$x^y=3/2$$ $$(x ^y)^{1/y}=\left(\frac{3}{2}\right)^{1/y}$$ $$x=\left(\frac{3}{2}\right)^{1/y}$$

However, the answer in the book is $\frac{3^{1/y}}{2}$. I think I know my mistake but I am not sure why it is wrong. I believe that my mistake was when I divided $2$ from $2x^y$, are you not allowed to do that?

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The mistake is in the first line. You should have $(2x)^y$ instead of $2(x^y)$. $(2x)$ is the base of the logarithm and so the entire term needs to be taken to the power of $y$.
Since the entire term is is raised to the power of $y$, I would have to take care of the $y$ first before dividing by 2 correct? I think I understand now. – Kot Oct 4 '12 at 23:36
Exactly! You don't have $2$ alone in the equation but rather a factor of $2^y$. You not only need to take care of the $y$ atop the $x$ but also the $2$. – EuYu Oct 4 '12 at 23:40
Well, you almost got it). If $y=f(x)$ then $x=f^{-1}(y)$.
$$y=f(x)=\log_{2x}3\\ (2x)^y=3\\ 2x=3^{\frac{1}{y}}\\ x=\frac{1}{2}3^{\frac{1}{y}}=f^{-1}(y)$$