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How would I determine $\left(\frac{9}{16}\right)^{\frac{3}{2}}$? I've never encountered an exponent that was a fraction, or a least never without a calculator. What steps would I take to simplify this?

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Recall a few of the properties of exponents:

$$\large a^{bc} = (a^b)^c\tag{1}$$ $$\large\left(\dfrac ab\right)^c = \frac{a^{\,c}}{b^{\,c}}\tag{2}$$ And for exponents of the form $\frac 1n$ where $n$ is a non-zero integer, for any non-negative $a$, or for all $a$ provided $n$ is odd, then $$\large a^{1/n} = \sqrt[\Large n]a \tag{3}$$ Using these properties:

$$\begin{align} \left(\frac 9{16}\right)^{3/2} & = \left(\left(\frac 9{16}\right)^{1/2}\right)^3 \tag{1} \\ \\ & = \left(\frac{(9)^{1/2}}{(16)^{1/2}}\right)^3 \tag{2}\\ \\ & = \left(\frac{\sqrt 9}{\sqrt{16}}\right)^3 \tag{3} \\ \\ & = \left(\frac 34\right)^3 = \frac{27}{64}\tag{evaluate}\end{align}$$

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helpful to show def - then answer +1 – Amzoti Jul 2 '13 at 2:53
It is definitely helpful for the OP. – Babak S. Jul 2 '13 at 5:21

Hint: Notice that: $$ \left( \dfrac{9}{16} \right)^{\frac{3}{2}} = \left(\left( \dfrac{3}{4} \right)^2\right)^{\frac{3}{2}}= \left( \dfrac{3}{4} \right)^{2 \cdot \frac{3}{2}}= \left( \dfrac{3}{4} \right)^{3} $$

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