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I just don't know what to do when I get to $4096$. The original equation was $16^{3/4}$.

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$$16^{3/4} = (2^4)^{3/4} = 2^{4\cdot(3/4)} = 2^3 = 8.$$

First we note that $16 = 2^4$, after that we use that $(a^n)^m = a^{n\cdot m}$ (this holds in general for all non-negative real $a$).

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How about $$16^{3/4} = [(-2)^4]^{3/4} = (-2)^{4\cdot(3/4)} = (-2)^3 = -8$$ – lab bhattacharjee May 18 '14 at 16:01
@labbhattacharjee I have clarified. – Eff May 18 '14 at 16:03
I guess the main point is you first do $\mbox{..}^{1/4}$ and then cube it. I would refrain from answering in full, though. – PA6OTA May 18 '14 at 16:05
@user112061, What's is wrong with my calculation in the comment? – lab bhattacharjee May 18 '14 at 16:06
@user112061, sort of seconds your point. – lab bhattacharjee May 18 '14 at 16:33

Let $\displaystyle a^4=16\implies a^2=\pm4$

$\displaystyle\implies a=\pm2,\pm2i$

the principal value of $a$ being $2$


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Fine answer, but a person asking how to compute $16^{3/4}$ may never have heard of $i$, even less of "principal values." – Eff May 18 '14 at 16:19
@user112061, Are these terms prohibited for him? In many a Questions, we are introduced with new formulas & terms, right? – lab bhattacharjee May 18 '14 at 16:22
Yes, but it has to be within a certain reach of what the person already knows to not further confuse him or her. Don't get me wrong, I thought your answer was good. – Eff May 18 '14 at 16:33

The really basic way to do this:

$$\begin{align}(16^3)^{1/4} &= ((2^4)^3)^{1/4}\\ &=2^{4 \cdot 3 \cdot 1/4}\\ &=2^3\\ &= 8 \end{align}$$

Note that -8 is an equally valid solution since the power is even. If the answer is meant to be "easy" then usually the solution is found by expressing things in terms of the prime factors.

As an aside, it gets more interesting when the answer is not an integer (in other words, when the number you start with is not a perfect square). There is a neat trick (which they no longer teach in most schools) for getting the square root of any number to arbitrary precision (limited by your patience, accuracy, size of your paper and the sharpness of your pencil). It looks a bit like long division. See "Method 2" of

It can work for any root that is a power of 2 (2, 4, 8...) - you just have to take the root of the root, etc.

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If you recognize that $16^{\frac{1}{4}}=2$ (since $2^4=2\cdot2\cdot2\cdot2=4\cdot4=16$), then $$16^{3/4}=(16^{\frac{1}{4}})^{3}=(2)^3=\boxed{8}$$

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Have you read the other answers? – Servaes May 18 '14 at 16:09

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