# How do you evaluate (7^ { 3/ 4} ) ^ { - 2/ 9}?

Apr 14, 2018

$\implies {\left(7\right)}^{- \frac{1}{6}}$

#### Explanation:

As per Laws of Indices,

(a^m)^n) = a ^(m * n)

${\left({7}^{\frac{3}{4}}\right)}^{-} \left(\frac{2}{9}\right) = {\left(7\right)}^{\left(\frac{3}{4}\right) \cdot \left(- \frac{2}{9}\right)}$

$\implies {\left(7\right)}^{- \frac{6}{36}} = {\left(7\right)}^{- \frac{1}{6}}$

${\left(7\right)}^{- \frac{1}{6}}$ can be rewritten as ${.}^{6} \sqrt{\frac{1}{7}}$.

I hope that helps!