Solving $x^{2n} = \frac{1}{2^n}$ for $x$ What is the principle behind solving for a variable that is raised to another variable?  I came across this problem doing infinite sums: I had to solve the equation 
$$x^{2n} = \frac{1}{2^n}$$
for $x$. I posed the question in the online forum and the TA said the answer is 
$$x = \frac{1}{\sqrt{2}}.$$  
I don't see how he got there.  If someone could explain how to get from one to the other, I would appreciate it!
 A: $x^{2n}=(x^2)^n=\frac{1}{2^n}$. So if $n\neq 0$, for $x\in \mathbb{R}$ you have $x^2=\frac{1}{2}$ then $x=\frac{1}{\sqrt{2}}$ or $x=-\frac{1}{\sqrt{2}}$.
A: To solve an expression like $x^{2n}=\frac{1}{2n}$ for $x$, you need to use logarithms. As was quite rightly pointed out, if $n\neq 0$ and $x\in\mathbb{R},$ then 
$$(x^2)^n=\frac{1}{2^n}\Rightarrow x^2=\frac{1}{2}\Rightarrow x=\frac{1}{\sqrt{2}}.$$ 
If $n=0$, then we have a blow-up singularity. Hence we will need to travel in an arbitrarily small circuit around the singularity (hence this extra term). 
Taking logarithms of both sides and using the log power rule gives
$$
2n\ln(x)=2\pi ik+\ln(\frac{1}{2n}),~k\in\mathbb{Z}.
$$
Travelling in an anticlockwise traversal along a simple closed curve around the origin gives the extra $2\pi ik$ term (the logarithm is given by $2\pi k$ multiplied by $k\in\mathbb{Z}$, with $k$ the winding number). . Dividing both sides by $2n$ gives
$$
\ln(x)=\frac{\pi ik}{n}+\frac{\ln(\frac{1}{2n})}{2n},~k\in\mathbb{Z}.
$$
Take the exponential function of both sides, and you have an expression for $x$.
