# How do you simplify (4 times 10^13)^-2?

Dec 7, 2015

$\left(\frac{1}{16}\right) \left(\frac{1}{10} ^ - 26\right) = \left(6.25 \cdot {10}^{-} 28\right)$
${\left({a}^{1} \cdot {b}^{x}\right)}^{n} = {a}^{1 n} \cdot {b}^{x n}$
In this case $4$ is raised to the power of one, so it becomes ${4}^{-} 2$, or $\frac{1}{16}$.
$10$ is raised to the power of $13$, so it becomes ${10}^{13 \cdot \left(- 2\right)}$ or ${10}^{-} 26$