I am trying to simplify this but I'm not sure how to approach it.
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$\begingroup$ $\left( \frac{a}{b}\right)^{\frac{c}{d}} = \frac{a^{\frac{c}{d}}}{b^{\frac{c}{d}}}$ $\endgroup$– quapkaNov 23, 2015 at 20:05
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5 Answers
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Write $\frac{1}{9} = 3^{-2}$.
Then
$\big(\frac{1}{9}\big)^{\frac{3}{2}} = 3^{-2 \cdot \frac{3}{2}} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27}$.
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Try applying logarithms:
$a^b = c \implies b \ln a = \ln c$
and you also know that $1^x=1$.
Good luck!
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$\begingroup$ just a dumb question why not use log base 10 ? $\endgroup$ Nov 28, 2018 at 20:18
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Hint: this is equivalent to $\sqrt{\left(\frac{1}{9}\right)^3}=\sqrt{\frac{1}{729}}$
answered Nov 23, 2015 at 20:02
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$(\frac{1}{9})^{\frac{3}{2}}=(\sqrt{\frac{1}{9}})^3=\frac{1}{27}$
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In general, if $a>0$, the following will hold:
$$a^{\frac pq}=(\sqrt[q]{a})^p.$$
In your case:
$$(\frac19)^{\frac32}=(\sqrt[2]{\frac19})^3=(\frac13)^3=\frac{1}{27}.$$
What about $a<0$? Well, if the power is defined, the above will hold.
