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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!

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    $\begingroup$ Please format your question better. Is it $x^2 n$ or $x^{2n}$? $\endgroup$ – user223391 Aug 22 '15 at 14:54
  • $\begingroup$ Similarly, is it $\frac{1}{2}n$, $\frac{1}{2n}$ or $\frac{1}{2^n}$? $\endgroup$ – user2103480 Aug 22 '15 at 15:06
  • $\begingroup$ It looks like x^(2n)=1/(2n). $\endgroup$ – Jonathan Aug 22 '15 at 15:08
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$x^{2n}=(x^2)^n=\frac{1}{2^n}$. So if $n\neq 0$, for $x\in \mathbb{R}$ you have $x^2=\frac{1}{2}$ then $x=\frac{1}{\sqrt{2}}$ or $x=-\frac{1}{\sqrt{2}}$.

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    $\begingroup$ what is if $$n=0$$? $\endgroup$ – Dr. Sonnhard Graubner Aug 22 '15 at 15:23
  • $\begingroup$ That is usually a special case. $\endgroup$ – marty cohen Aug 22 '15 at 16:54
  • $\begingroup$ Alternatively $$x^{2n}=\frac{1}{2^n}=\bigg(\frac{1}{2}\bigg)^n=\bigg(\frac{1}{2}\bigg)^\frac{2n}{2}=\bigg(\bigg(\frac{1}{2}\bigg)^\frac{1}{2}\bigg)^{2n}=\bigg(\frac{1}{\sqrt{2}}\bigg)^{2n}\implies\text{ yada yada yada}$$ $\endgroup$ – John Joy Aug 22 '15 at 17:40
  • $\begingroup$ How would you solve for x if the question had been $ x^2n=1/2n $? $\endgroup$ – Jonathan Aug 22 '15 at 22:21
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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 $x$.

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  • $\begingroup$ Your reputation at this moment corresponds to the first four digits of $\pi$ (rounded to 3dp)! $\endgroup$ – hypergeometric Aug 22 '15 at 17:07
  • $\begingroup$ So it does! Fancy that! $\endgroup$ – user230715 Aug 22 '15 at 17:09
  • $\begingroup$ Better take a screenshot before it changes :) $\endgroup$ – hypergeometric Aug 22 '15 at 17:09
  • $\begingroup$ Done that already! $\endgroup$ – user230715 Aug 23 '15 at 9:41
  • $\begingroup$ can you tell me why you added this 2πik $\endgroup$ – RajSharma Aug 31 '15 at 9:15

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