Question

rewrite the following without an exponent: $\left(\frac{9}{8}\
ight)^{-2}$

Explanation:

Step1: Apply negative - exponent rule

According to the rule $a^{-n}=\frac{1}{a^{n}}$, for $a = \frac{9}{8}$ and $n = 2$, we have $(\frac{9}{8})^{-2}=\frac{1}{(\frac{9}{8})^{2}}$.

Step2: Square the fraction

$(\frac{9}{8})^{2}=\frac{9^{2}}{8^{2}}=\frac{81}{64}$. So $\frac{1}{(\frac{9}{8})^{2}}=\frac{64}{81}$.

Answer:

$\frac{64}{81}$