Fractional exponents I know that $64^{\frac{1}{3}}$ is basically the cube root, but what if it's $64^{\frac{2}{3}}$ ?
Like, how would that look like numerically? 
Any help is much appreciated, thanks.
 A: Remember that $\frac{a}{b}=a \times \frac{1}{b}$. Therefore,
$$64^{\frac{2}{3}}=64^{2 \times \frac{1}{3}}=(64^{2})^{\frac{1}{3}}$$
A: The numerator of the power is what you raise your number to, and the denominator is the root to which you take the number.
The order (which operation you do first) does not matter!
Note:
4^(3/2) = 4^3*(1/2)
4^(1/2) = 2
2^3 = 8
or
4^3 = 64
64^(1/2) = 8
Either way, you get the same result.
The key is:
x^(a/b) = (x^a)^(1/b) = (x^(1/b))^a
In your specific case:
64^(2/3) = (64^(1/3))^2 = (4)^2 = 16
If you are looking for intuition: when you exponentiate a number to a fraction,
we are essentially saying:
if x^(a/b), then take the "bth" (e.g. 4th, 2nd, 7th, etc.) of x and multiply it by itself "a" many times.
Hence take: x^(a/a) = x = "ath" roth mutiplied by itself a times, which gives back x 
If you would like more explanations, please leave a comment :)!!
A: Well $64^{1\over3}=\sqrt[3]{64}$, and $64=64^1$, therefore $64^{1\over3}=\sqrt[3]{64^1}$.
By this logic $$64^{2\over3}=\sqrt[3]{64^2}$$
