# Why does $16^{1/3} = 2^{4/3}$

I'm working through an example problem in my text book and they simplify $16^{1/3}$ to $2^{4/3}$. They also simplify $\frac{1}{2}(2^{8/3})$ to $2^{5/3}$. I don't follow the logic.

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$$16^{1/3} = (2^4)^{1/3} = 2^{4\cdot(1/3)}.$$

$$\frac12(2^{8/3}) = 2^{-1}\cdot 2^{8/3} = 2^{-1+(8/3)}$$

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A short and simple answer:

You just need to know that $\sqrt[a]{x^b} = x^{\large{b \over a}}$ and $a^n \div a^f = a^{n - f}$ .

Now,$$16^{\large{1 \over 3}} = \sqrt[3]{16} = \sqrt[3]{2^4} = 2^{\large{4 \over 3}}$$

The second, $${1 \over 2}\cdot 2^{8 \over 3} = 2^{8 \over 3} \div 2^1 = 2^{{8 \over 3} - 1}=2^{5 \over 3}$$ Ta-da!

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I wish SE would have a "like" option alongside the upvote. –  akkkk Nov 18 '12 at 14:57