Evaluate $\frac{3}{4}$ to the power of $-3$

Question

I have gotten to the next stage where you write it as $\frac{1}{\left(\frac 34\right)}$ to the power of $3$, now I am stuck

I've got it now, thanks everyone.

You should use the relation $\frac{1}{\frac{a}{b}}=\frac{b}{a}$. Can you continue from here? — Tintarn, Sep 23 '15 at 17:19
@rooroo, I made a small edit that might help with your confusion — jameselmore, Sep 23 '15 at 17:31

jameselmore · Answer 1 · 2015-09-23T17:31:34.207

5

Hint: $$\left(\frac{b}{a}\right)^{-n} = \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$$

edited Sep 23 '15 at 17:31

answered Sep 23 '15 at 17:19

jameselmore

score 2 · Answer 2 · edited Sep 23 '15 at 17:51

2

$$=\left(\frac{3}{4}\right)^{-3}=\left(\frac{1}{\frac{3}{4}}\right)^3=\left(\frac{4}{3}\right)^3=\frac{4^3}{3^3}=\frac{64}{27}\approx2.37$$

edited Sep 23 '15 at 17:51

user35603

answered Sep 23 '15 at 17:21

Mark

score 0 · Answer 3 · answered Sep 23 '15 at 17:22

0

it is $(\frac{4}{3})^3=\frac{4^3}{3^3}=\frac{64}{27}$

answered Sep 23 '15 at 17:22

So just to clarify, I need to swap the 3 and 4 over? – rooroo Sep 23 '15 at 17:30
yes this is right – Dr. Sonnhard Graubner Sep 23 '15 at 17:31
and not make it 1 over anything? – rooroo Sep 23 '15 at 17:31
it is $$\left(\frac{3}{4}\right)^{-3}=\left(\frac{4}{3}\right)^3$$ – Dr. Sonnhard Graubner Sep 23 '15 at 17:33

3 Answers3