# 0.2 Essential mathematics

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## Exponential arithmetic

Exponential notation is used to express very large and very small numbers as a product of two numbers. The first number of the product, the digit term , is usually a number not less than 1 and not greater than 10. The second number of the product, the exponential term , is written as 10 with an exponent. Some examples of exponential notation are:

$\begin{array}{ccc}\hfill 1000& =& 1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\hfill \\ \hfill 100& =& 1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\hfill \\ \hfill 10& =& 1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{1}\hfill \\ \hfill 1& =& 1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{0}\hfill \\ \hfill 0.1& =& 1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}\hfill \\ \hfill 0.001& =& 1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\hfill \\ \hfill 2386& =& 2.386\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}1000=2.386\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\hfill \\ \hfill 0.123& =& 1.23\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.1=1.23\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}\hfill \end{array}$

The power (exponent) of 10 is equal to the number of places the decimal is shifted to give the digit number. The exponential method is particularly useful notation for every large and very small numbers. For example, 1,230,000,000 = 1.23 $×$ 10 9 , and 0.00000000036 = 3.6 $×$ 10 −10 .

Convert all numbers to the same power of 10, add the digit terms of the numbers, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.

Add 5.00 $×$ 10 −5 and 3.00 $×$ 10 −3 .

## Solution

$\begin{array}{ccc}\hfill 3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}& =& 300\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\hfill \\ \hfill \left(5.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\right)+\left(300\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\right)& =& 305\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}=3.05\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\hfill \end{array}$

## Subtraction of exponentials

Convert all numbers to the same power of 10, take the difference of the digit terms, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term.

## Subtracting exponentials

Subtract 4.0 $×$ 10 −7 from 5.0 $×$ 10 −6 .

## Solution

$\begin{array}{}\\ 4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}=0.40\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\\ \left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)-\left(0.40\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)=4.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\end{array}$

## Multiplication of exponentials

Multiply the digit terms in the usual way and add the exponents of the exponential terms.

## Multiplying exponentials

Multiply 4.2 $×$ 10 −8 by 2.0 $×$ 10 3 .

## Solution

$\left(4.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\right)=\left(4.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2.0\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\left(-8\right)+\left(+3\right)}=8.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

## Division of exponentials

Divide the digit term of the numerator by the digit term of the denominator and subtract the exponents of the exponential terms.

## Dividing exponentials

Divide 3.6 $×$ 10 5 by 6.0 $×$ 10 −4 .

## Solution

$\frac{3.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}}{6.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}}\phantom{\rule{0.2em}{0ex}}=\left(\frac{3.6}{6.0}\right)\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\left(-5\right)-\left(-4\right)}=0.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-1}=6.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}$

## Squaring of exponentials

Square the digit term in the usual way and multiply the exponent of the exponential term by 2.

## Squaring exponentials

Square the number 4.0 $×$ 10 −6 .

## Solution

${\left(4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)}^{2}=4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(-6\right)}=16\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-12}=1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}$

## Cubing of exponentials

Cube the digit term in the usual way and multiply the exponent of the exponential term by 3.

## Cubing exponentials

Cube the number 2 $×$ 10 4 .

## Solution

${\left(2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\right)}^{3}=2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}4}=8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{12}$

## Taking square roots of exponentials

If necessary, decrease or increase the exponential term so that the power of 10 is evenly divisible by 2. Extract the square root of the digit term and divide the exponential term by 2.

## Finding the square root of exponentials

Find the square root of 1.6 $×$ 10 −7 .

## Solution

$\begin{array}{ccc}\hfill 1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}& =& 16\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\hfill \\ \hfill \sqrt{16\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}}=\sqrt{16}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\sqrt{{10}^{-8}}& =\hfill & \sqrt{16}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-\phantom{\rule{0.2em}{0ex}}\frac{8}{2}}\phantom{\rule{0.2em}{0ex}}=4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\hfill \end{array}$

## Significant figures

A beekeeper reports that he has 525,341 bees. The last three figures of the number are obviously inaccurate, for during the time the keeper was counting the bees, some of them died and others hatched; this makes it quite difficult to determine the exact number of bees. It would have been more accurate if the beekeeper had reported the number 525,000. In other words, the last three figures are not significant, except to set the position of the decimal point. Their exact values have no meaning useful in this situation. In reporting any information as numbers, use only as many significant figures as the accuracy of the measurement warrants.

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