 # How to Calculate with Scientific Notation

Here, you will take a look at how you do calculations with numbers in scientific notations. It’s a piece of cake. The main idea is to work separately with the power of 10’s and the numbers in front of the powers.

Rule

### MultiplicationwithScientificNotations

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(a\cdot 1{0}^{n}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(b\cdot 1{0}^{m}\right)& =a\cdot 1{0}^{n}\cdot b\cdot 1{0}^{m}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =a\cdot b\cdot 1{0}^{n+m}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\phantom{\rule{-0.17em}{0ex}}\left(a\cdot 1{0}^{n}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(b\cdot 1{0}^{m}\right)=a\cdot 1{0}^{n}\cdot b\cdot 1{0}^{m}=a\cdot b\cdot 1{0}^{n+m}$

Rule

### DivisionwithScientificNotation

 $\frac{a\cdot 1{0}^{n}}{b\cdot 1{0}^{m}}=\frac{a}{b}\cdot \frac{1{0}^{n}}{1{0}^{m}}=\frac{a}{b}\cdot 1{0}^{n-m}$

Example 1

Calculate $\phantom{\rule{-0.17em}{0ex}}\left(4\cdot 1{0}^{4}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(2\cdot 1{0}^{2}\right)$

You use the multiplication rule.

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(4\cdot 1{0}^{4}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(2\cdot 1{0}^{2}\right)& =4\cdot 2\cdot 1{0}^{4+2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8\cdot 1{0}^{6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\phantom{\rule{-0.17em}{0ex}}\left(4\cdot 1{0}^{4}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(2\cdot 1{0}^{2}\right)=4\cdot 2\cdot 1{0}^{4+2}=8\cdot 1{0}^{6}$

Example 2

Calculate $5\cdot 1{0}^{4}\cdot 3\cdot 1{0}^{5}$

Remember to sort. $\begin{array}{llll}\hfill 5\cdot 1{0}^{4}\cdot 3\cdot 1{0}^{5}& =5\cdot 3\cdot 1{0}^{4+5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =15\cdot 1{0}^{9}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1.5\cdot 10\cdot 1{0}^{9}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1.5\cdot 1{0}^{9+1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1.5\cdot 1{0}^{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

A value in scientific notation has a number from $1$ to $10$ multiplied by a power of 10. Since $5\cdot 3=$15, you will need to convert $15$ to a number between $1$ and $10$. To do this, write $15=1.5\cdot 10$ (in scientific notations) and repeat once more.

Example 3

Calculate $\frac{8\cdot 1{0}^{6}}{4\cdot 1{0}^{4}}$

You use the division rule.

$\begin{array}{llll}\hfill \frac{8\cdot 1{0}^{6}}{4\cdot 1{0}^{4}}& =\frac{8}{4}\cdot \frac{1{0}^{6}}{1{0}^{4}}=2\cdot 1{0}^{6-4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\cdot 1{0}^{2}=200\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\frac{8\cdot 1{0}^{6}}{4\cdot 1{0}^{4}}=\frac{8}{4}\cdot \frac{1{0}^{6}}{1{0}^{4}}=2\cdot 1{0}^{6-4}=2\cdot 1{0}^{2}=200$

Example 4

Calculate $\frac{3\cdot 1{0}^{9}}{6\cdot 1{0}^{5}}$ and write in scientific notation

You factorize into two fractions $\begin{array}{llll}\hfill \frac{3\cdot 1{0}^{9}}{6\cdot 1{0}^{5}}& =\frac{3}{6}\cdot \frac{1{0}^{9}}{1{0}^{5}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.5\cdot 1{0}^{9-5}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.5\cdot 1{0}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5\cdot 1{0}^{-1}\cdot 1{0}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5\cdot 1{0}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

A value in scientific notation has a number from $1$ to $10$ multiplied by a power of 10. You must convert $0.5$ to a number between $1$ and $10$. To do this, write $0.5=5\cdot 1{0}^{-1}$ (in scientific notation) and repeat once more. 