Question

Use the law of exponents to rewrite and simplify the expression.

$\frac{4^{-3}}{2^{-8}}$

Answer 1

To simplify the given expression:

Using the law of exponents to simplify the given expression in below:

$\begin{array}{rcl}\frac{4^{-3}}{2^{-8}}& =& \frac{{\left(2^2\right)}^{-3}}{2^{-8}}\left[\text{Rewrite}4\text{as}{2}^2\right]\\ & \Rightarrow & \frac{4^{-3}}{2^{-8}}=\frac{2^{2\times -3}}{2^{-8}}\left[\text{Power of a power rule}:{\left({\text{a}}^{\text{m}}\right)}^{\text{n}}={\text{a}}^{\text{m}\times \text{n}}\right]\\ & \Rightarrow & \frac{4^{-3}}{2^{-8}}=\frac{2^{-6}}{2^{-8}}\left[\text{Simplifying}\right]\\ & \Rightarrow & \frac{4^{-3}}{2^{-8}}=2^{-6+8}\left[\text{Quotient rule}:\frac{{\text{a}}^{\text{m}}}{{\text{a}}^{\text{n}}}={\text{a}}^{\text{m}-\text{n}}\right]\\ & \Rightarrow & \frac{4^{-3}}{2^{-8}}=2^2\left[\text{Simplifying}\right]\\ \therefore \frac{4^{-3}}{2^{-8}}& =& 4\end{array}$

Hence, the answer is $\mathbf{4}$.