How to raise a fraction to a fractional exponent?

Question

I am trying to simplify this but I'm not sure how to approach it.

$\left( \frac{a}{b}\right)^{\frac{c}{d}} = \frac{a^{\frac{c}{d}}}{b^{\frac{c}{d}}}$ — quapka, Nov 23 '15 at 20:05

score 6 · Accepted Answer · answered Nov 23 '15 at 20:01

6

Write $\frac{1}{9} = 3^{-2}$.

Then

$\big(\frac{1}{9}\big)^{\frac{3}{2}} = 3^{-2 \cdot \frac{3}{2}} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27}$.

answered Nov 23 '15 at 20:01

Tien Truong

score 2 · Answer 2 · edited Nov 23 '15 at 20:13

2

Try applying logarithms:

$a^b = c \implies b \ln a = \ln c$

and you also know that $1^x=1$.

Good luck!

edited Nov 23 '15 at 20:13

type_outcast

answered Nov 23 '15 at 20:00

Kio

score 0 · Answer 3 · answered Nov 23 '15 at 20:02

0

Hint: this is equivalent to $\sqrt{\left(\frac{1}{9}\right)^3}=\sqrt{\frac{1}{729}}$

answered Nov 23 '15 at 20:02

score 0 · Answer 4 · answered Nov 23 '15 at 20:06

0

$(\frac{1}{9})^{\frac{3}{2}}=(\sqrt{\frac{1}{9}})^3=\frac{1}{27}$

answered Nov 23 '15 at 20:06

Erfan Khaniki

score 0 · Answer 5 · answered Nov 23 '15 at 20:40

0

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.

answered Nov 23 '15 at 20:40

MickG

5 Answers5