# How do you simplify 9times10^ -31 * (3times10 ^7)^2 ?

Nov 9, 2015

$81 \cdot {10}^{- 17}$

#### Explanation:

First of all, simplify the square in the second part:

$9 \cdot {10}^{- 31} \cdot {\left(3 \cdot {10}^{7}\right)}^{2} = 9 \cdot {10}^{- 31} \cdot {3}^{2} \cdot {\left({10}^{7}\right)}^{2}$

According to the power rule ${\left({a}^{n}\right)}^{m} = {a}^{n \cdot m}$, you can simplify further:

$\ldots = 9 \cdot {10}^{- 31} \cdot 9 \cdot {10}^{14}$

Now, compute $9 \cdot 9$:

$\ldots = 81 \cdot {10}^{- 31} \cdot {10}^{14}$

Use the power rule: ${a}^{n} \cdot {a}^{m} = {a}^{n + m}$

$\ldots = 81 \cdot {10}^{- 31 + 14} = 81 \cdot {10}^{- 17}$

If you would like to formulate the result in the scientific notation, you would need to shift one digit of the $81$ and thus increase the exponent of the basis $10$ by $1$:
$\ldots = 8.1 \cdot {10}^{- 16}$.