# 3.7 Scientific notation  (Page 2/3)

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$\text{}0.000000000004632\text{}=\text{}4.632×{10}^{-12}$

The decimal point is twelve places to the right of its original position, and the power of 10 is $-12$ .

$0.027=\text{\hspace{0.17em}}2.7×{10}^{-2}$

The decimal point is two places to the right of its original position, and the power of 10 is $-2$ .

## Practice set a

Write the following numbers in scientific notation.

346

$3.46×{10}^{2}$

$72.33$

$7.233×10$

$5387.7965$

$5.3877965×{10}^{3}$

87,000,000

$8.7×{10}^{7}$

179,000,000,000,000,000,000

$1.79×{10}^{20}$

100,000

$1.0×{10}^{5}$

1,000,000

$1.0×{10}^{6}$

$0.0086$

$8.6×{10}^{-3}$

$0.000098001$

$9.8001×{10}^{-5}$

$0.000000000000000054$

$5.4×{10}^{-17}$

$0.0000001$

$1.0×{10}^{-7}$

$0.00000001$

$1.0×{10}^{-8}$

## Scientific form to standard form

A number written in scientific notation can be converted to standard form by reversing the process shown in Sample Set A.

## Converting from scientific notation

To convert a number written in scientific notation to a number in standard form, move the decimal point the number of places prescribed by the exponent on the 10.

## Positive exponent negative exponent

Move the decimal point to the right when you have a positive exponent, and move the decimal point to the left when you have a negative exponent.

## Sample set b

$4.673×{10}^{4}$ .

The exponent of 10 is 4 so we must move the decimal point to the right 4 places (adding $0\text{'}\text{s}$ if necessary).

$2.9×{10}^{7}$ .

The exponent of 10 is 7 so we must move the decimal point to the right 7 places (adding $0\text{'}\text{s}$ if necessary).

$2.9×{10}^{7}=\text{29000000}$

$1×{10}^{27}$ .

The exponent of 10 is 27 so we must move the decimal point to the right 27 places (adding $0\text{'}\text{s}$ without a doubt).

$1×{10}^{27}=\text{1,000,000,000,000,000,000,000,000,000}$

$4.21×{10}^{-5}$ .

The exponent of 10 is $-5$ so we must move the decimal point to the left 5 places (adding $0\text{'}\text{s}$ if necessary).

$4.21×{10}^{-5}=\text{0}\text{.0000421}$

$1.006×{10}^{-18}$ .

The exponent of 10 is $-18$ so we must move the decimal point to the left 18 places (adding $0\text{'}\text{s}$ if necessary).

$1.006×{10}^{-18}=0.000000000000000001006$

## Practice set b

Convert the following numbers to standard form.

$9.25×{10}^{2}$

925

$4.01×{10}^{5}$

401000

$1.2×{10}^{-1}$

$0.12$

$8.88×{10}^{-5}$

$0.0000888$

## Multiplying numbers using scientific notation

There are many occasions (particularly in the sciences) when it is necessary to find the product of two numbers written in scientific notation. This is accomplished by using two of the basic rules of algebra.

Suppose we wish to find $\left(a×{10}^{n}\right)\left(b×{10}^{m}\right)$ . Since the only operation is multiplication, we can use the commutative property of multiplication to rearrange the numbers.

$\left(a×{10}^{n}\right)\left(b×{10}^{m}\right)=\left(a×b\right)\left({10}^{n}×{10}^{m}\right)$

Then, by the rules of exponents, ${10}^{n}×{10}^{m}={10}^{n+m}$ . Thus,

$\left(a×{10}^{n}\right)\left(b×{10}^{m}\right)=\left(a×b\right)×{10}^{n+m}$

The product of $\left(a×b\right)$ may not be between 1 and 10, so $\left(a×b\right)×{10}^{n+m}$ may not be in scientific form. The decimal point in $\left(a×b\right)$ may have to be moved. An example of this situation is in Sample Set C, problem 2.

## Sample set c

$\begin{array}{ll}\left(2×{10}^{3}\right)\left(4×{10}^{8}\right)\hfill & =\left(2×4\right)\left({10}^{3}×{10}^{8}\right)\hfill \\ \hfill & =8×{10}^{3+8}\hfill \\ \hfill & =8×{10}^{11}\hfill \end{array}$

$\begin{array}{ll}\left(5×{10}^{17}\right)\left(8.1×{10}^{-22}\right)\hfill & =\left(5×8.1\right)\left({10}^{17}×{10}^{-22}\right)\hfill \\ \hfill & =40.5×{10}^{17-22}\hfill \\ \hfill & =40.5×{10}^{-5}\hfill \end{array}$

We need to move the decimal point one place to the left to put this number in scientific notation.

Thus, we must also change the exponent of 10.

$\begin{array}{l}40.5×{10}^{-5}\hfill \\ 4.05×{10}^{1}×{10}^{-5}\hfill \\ 4.05×\left({10}^{1}×{10}^{-5}\right)\hfill \\ 4.05×\left({10}^{1-5}\right)\hfill \\ 4.05×{10}^{-4}\hfill \end{array}$

Thus,

$\left(5×{10}^{17}\right)\left(8.1×{10}^{-22}\right)=4.05×{10}^{-4}$

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