# How do you divide (3.0 x 10^4)/ (7 x 10^-7) ?

Mar 29, 2015

I believe the question is about simplifying the number written as
$\frac{3.0 \cdot {10}^{4}}{7 \cdot {10}^{-} 7}$
(it's a sign of multiplication, not a variable $x$ in the numerator and denominator).

The answer is (approximately) $0.428571 \cdot {10}^{11}$.
It is an approximate value because a rational number $\frac{3}{7}$ cannot be represented as a finite decimal fraction, it's "precise" value is a periodical decimal fraction with $\left(428571\right)$ being a period. So, the precise decimal representation of the original number is $0. \left(428571\right) \cdot {10}^{11}$, where the parenthesis are used to specify a period.

It's all about exponents. Negative exponent is used as a replacement for multiplicative inverse numbers (multiplicative inverse of $N$ is $\frac{1}{N}$).
Thus, ${10}^{-} 7$ is a replacement for $\frac{1}{10} ^ 7$.

It's not just a syntactical convenience, there is a very good mathematical reason for this representation based on the rules of multiplication of numbers expressed as exponents with the same base:
${A}^{m} \cdot {A}^{n} = {A}^{m + n}$ and
${A}^{0} = 1$
You can learn the details of how to deal with exponential functions at Unizor by following the menu items Algebra and Exponential Functions.

Using the above mention rule about negative exponents, you can replace ${10}^{-} 7$ in the denominator by $\frac{1}{10} ^ 7$, which will cause having both ${10}^{4}$ and ${10}^{7}$ in the numerator, so the expression will look like
$\frac{3.0 \cdot {10}^{4} \cdot {10}^{7}}{7}$,
which, in turn, using the rule of multiplication of exponential numbers with the same base $10$, can be represented as
$\frac{3}{7} \cdot {10}^{11}$.

All we have to do next is to represent $\frac{3}{7}$ as a decimal fraction, which we explained in the beginning, getting the final result
$0. \left(428571\right) \cdot {10}^{11}$,
where the parenthesis are used to specify a periodical decimal fraction.