# How do you divide (8.1*10^2)/(9.0*10^-3)?

Mar 29, 2015

You can deal separately with the numeric and the exponential part. So, first of all, divide $8.1$ by $9.0$, obtaining $0.9$. Then, for the exponential part, the exponent will be the sum of the powers at the numerator, minus the powers to the denominator. So, you have ${10}^{2 - \left(- 3\right)} = {10}^{2 + 3} = {10}^{5}$.

$\setminus \frac{8.1 \cdot {10}^{2}}{9.0 \cdot {10}^{- 3}} = 0.9 \cdot {10}^{5}$

Mar 29, 2015

Divide the coefficients and then the exponents separately before recombining.
$\frac{8.1 \cdot {10}^{2}}{9.0 \cdot 10 - 3}$

$= \frac{8.1}{9.0} \times \frac{{10}^{2}}{{10}^{- 3}}$

$= 0.9 \times \left({10}^{2} \cdot {10}^{3}\right)$

$= 0.9 \times {10}^{5}$

which would usually be normalized as:

$9.0 \times {10}^{4}$