How to solve for $m/n$ given that $(2x^{\frac{1}{6}})^{m} = (2x^{\frac{1}{8}})^{2n}$ for some constant $x$? I was solving a problem when I got stuck on a particular step, I don't know how to proceed further.
$$(2x^{\frac{1}{6}})^{m} = (2x^{\frac{1}{8}})^{2n}$$
I want to make the bases equal, so I can compare the exponents (i.e., the ratio of $m$ to $n$). Here $x$ is constant.
I tried but ended up making $x$ terms equal but then 2's exponents (in the base) became different.
Here is the original question:

If $m$ and $n$ are the smallest positive integers satisfying the
  relation $$\left(2\, \mathrm{cis} \frac\pi6\right)^m = \left(4\, \mathrm{cis} \frac\pi4\right)^n$$ then find the value of $m+n$.

where $\mathrm{cis}\, \theta \equiv \cos \theta + i \sin\theta$
 A: For the original problem
$$\left(2e^{i\frac{\pi}6}\right)^m=\left(4e^{i\frac{\pi}4}\right)^{n}$$
multiply both sides for
$$\left(\frac14e^{-i\frac{\pi}4}\right)^{n}$$
$$\left(2e^{i\frac{\pi}6}\right)^m\left(\frac14e^{-i\frac{\pi}4}\right)^{n}=1$$
$$2^{m-2n}e^{i\left(\frac{m\pi}6-\frac{n\pi}4\right)}=1$$
that is
$$\begin{cases}
m-2n=0\\
\frac{m\pi}6-\frac{n\pi}4=2k\pi
\end{cases}
\implies \frac{n}3-\frac{n}4=2k\implies n=24 \quad m=48 $$
A: Solving for $x$:
$$(2x^{1/6})^m=(2x^{1/8})^{2n}$$
$$2^mx^{m/6}=2^{2n}x^{2n/8}$$
$$x^{m/6}=2^{2n-m}x^{n/4}$$
$$x^{m/6-n/4}=2^{2n-m}$$
$$x=2^{\frac{2n-m}{m/6-n/4}}$$
$$x=2^{\frac{48n-24m}{4m-6n}}$$
$$x=2^{\frac{24n-12m}{2m-3n}}$$
$$x=2^{\frac{8*3n-6*2m}{2m-3n}}$$
$$x=2^{-\frac{6*2m-8*3n}{2m-3n}}$$
$$x=2^{-\frac{6*2m-6*3n-2*3n}{2m-3n}}$$
$$x=2^{\frac{6n}{2m-3n}-6}$$
$$x=\frac{1}{2^6}2^{\frac{6n}{2m-3n}}$$
The ratio of $m$ and $n$:
$$(2x^{1/6})^m=(2x^{1/8})^{2n}$$
$$m\log(2x^{1/6})=2n\log(2x^{1/8})$$
$$\frac{m}{n}=2\log_{2x^{1/6}}(2x^{1/8})$$
Or:
$$\frac{m}{n}=2\frac{\log(2)+\frac{1}{8}\log(x)}{\log(2)+\frac{1}{6}\log(x)}$$
$$\frac{m}{n}=2\frac{6\log(2)+\frac{6}{8}\log(x)}{6\log(2)+\log(x)}$$
$$\frac{m}{n}=2\frac{6\log(2)+(1-\frac{2}{8})\log(x)}{6\log(2)+\log(x)}$$
$$\frac{m}{n}=2\frac{6\log(2)+\log(x)-\frac{2}{8}\log(x)}{6\log(2)+\log(x)}$$
$$\frac{m}{n}=2\left(1-\frac{\frac{2}{8}\log(x)}{6\log(2)+\log(x)}\right)$$
$$\frac{m}{n}=2-\frac{1}{2}\frac{\log(x)}{6\log(2)+\log(x)}$$
The origial question:
$$2^m \text{cis}(\frac{\pi m}{6})=4^n \text{cis}(\frac{\pi n}{4})$$
$$2^{m-2n}=\text{cis}(\pi (\frac{n}{4}-\frac{m}{6}))$$
The left side is real, so the right side must be real aswell:
$$\sin(\pi (\frac{n}{4}-\frac{m}{6}))=0$$
$$\cos(\pi (\frac{n}{4}-\frac{m}{6}))=2^{m-2n}$$
For the first equation:
$$\pi (\frac{n}{4}-\frac{m}{6})=k \pi$$
$$\frac{n}{4}-\frac{m}{6}=k$$
$$6n-4m=24k$$
For the second, the right side must be 1:
$$m-2n=0$$
So:
$$6n-8n=24k$$
$$-2n=24k$$
$$n=12k$$
$$m=24k$$
But the $\cos$ must be positive, so the smallest possible $k$ value is $2$ (you get $-1$ as the value of the $\cos$ with $k=1$, so the next integer ($k=2$) will give $1$): $n=24$, $m=48$, so $n+m=72$.
