How do you solve for x in this radical equation?

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sqrt[x]{64} = 4

How do you solve for x?

I mean obviously the answer is x = 3 but how do you prove this algebraically?
 
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You can use logarithms or 'play' with powers:

\sqrt[x]{{64}} = 4 \Leftrightarrow 64^{1/x} = 4 \Leftrightarrow \left( {4^3 } \right)^{1/x} = 4 \Leftrightarrow 4^{3/x} = 4^1 \Leftrightarrow \frac{3}<br /> {x} = 1
 
I really, really wish people would say "x-root" rather than "x-square root" or root[x} instead of (as here) sqrt[x]. "square root" means specifically
^2\sqrt{x}
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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