# Question #1c88f

Sep 1, 2017

$1.2 \cdot {10}^{- 11}$

#### Explanation:

The thing to remember when multiplying numbers written in scientific notation is that you must multiply the mantissae and the exponents separately.

For a number written in scientific notation, you have

$\textcolor{w h i t e}{a a} \textcolor{b l u e}{m} \times {10}^{\textcolor{p u r p \le}{n} \textcolor{w h i t e}{a} \stackrel{\textcolor{w h i t e}{a a a a a a}}{\leftarrow}} \textcolor{w h i t e}{a \textcolor{b l a c k}{\text{the")acolor(purple)("exponent}} a a}$
$\textcolor{w h i t e}{\frac{a}{a} \textcolor{b l a c k}{\uparrow} a a a a}$
$\textcolor{w h i t e}{\textcolor{b l a c k}{\text{the")acolor(blue)("mantissa}} a}$

$\textcolor{b l u e}{3.0} \cdot {10}^{\textcolor{p u r p \le}{- 14}} \text{ " and " } \textcolor{b l u e}{4.0} \cdot {10}^{\textcolor{p u r p \le}{2}}$

This means that when you multiply these two numbers, you have

$\textcolor{b l u e}{3.0} \cdot {10}^{\textcolor{p u r p \le}{- 14}} \cdot \textcolor{b l u e}{4.0} \cdot {10}^{\textcolor{p u r p \le}{2}}$

$= \left(\textcolor{b l u e}{3.0 \cdot 4.0}\right) \cdot \left({10}^{\textcolor{p u r p \le}{- 14}} \cdot {10}^{\textcolor{p u r p \le}{2}}\right)$

$= \textcolor{b l u e}{12} \cdot {10}^{\textcolor{p u r p \le}{\left(- 14 + 2\right)}}$

$= \textcolor{b l u e}{12} \cdot {10}^{\textcolor{p u r p \le}{- 12}}$

More often than not, you will be dealing with normalized scientific notation, for which

$1 \le | \textcolor{b l u e}{m} | < 10$

To express the result of the multiplication in normalized scientific notation, divide the mantissa by $10$ and multiply by $10$ by adding $1$ to the exponent.

$\textcolor{b l u e}{12} \cdot {10}^{\textcolor{p u r p \le}{- 12}}$

$= \frac{\textcolor{b l u e}{12}}{10} \cdot {10}^{\textcolor{p u r p \le}{- 12}} \cdot 10$

$= \textcolor{b l u e}{1.2} \cdot {10}^{\textcolor{p u r p \le}{- 11}}$

Therefore, you can say that

$3.0 \cdot {10}^{- 14} \cdot 4.0 \cdot {10}^{2} = 1.2 \cdot {10}^{- 11}$

The answer is rounded to two sig figs, the number of sig figs you have for the two numbers.