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$\text{}0.000000000004632\text{}=\text{}4.632\times {10}^{-12}$
The decimal point is twelve places to the right of its original position, and the power of 10 is $-12$ .
$0.027=\text{\hspace{0.17em}}2.7\times {10}^{-2}$
The decimal point is two places to the right of its original position, and the power of 10 is $-2$ .
Write the following numbers in scientific notation.
A number written in scientific notation can be converted to standard form by reversing the process shown in Sample Set A.
$4.673\times {10}^{4}$ .
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).
$2.9\times {10}^{7}$ .
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).
$2.9\times {10}^{7}=\text{29000000}$
$1\times {10}^{27}$ .
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).
$1\times {10}^{27}=\text{1,000,000,000,000,000,000,000,000,000}$
$4.21\times {10}^{-5}$ .
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).
$4.21\times {10}^{-5}=\text{0}\text{.0000421}$
$1.006\times {10}^{-18}$ .
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).
$1.006\times {10}^{-18}=0.000000000000000001006$
Convert the following numbers to standard form.
Suppose we wish to find $(a\times {10}^{n})(b\times {10}^{m})$ . Since the only operation is multiplication, we can use the commutative property of multiplication to rearrange the numbers.
$(a\times {10}^{n})(b\times {10}^{m})=(a\times b)({10}^{n}\times {10}^{m})$
Then, by the rules of exponents, ${10}^{n}\times {10}^{m}={10}^{n+m}$ . Thus,
$(a\times {10}^{n})(b\times {10}^{m})=(a\times b)\times {10}^{n+m}$
The product of $(a\times b)$ may not be between 1 and 10, so $(a\times b)\times {10}^{n+m}$ may not be in scientific form. The decimal point in $(a\times b)$ may have to be moved. An example of this situation is in Sample Set C, problem 2.
$\begin{array}{ll}(2\times {10}^{3})(4\times {10}^{8})\hfill & =(2\times 4)({10}^{3}\times {10}^{8})\hfill \\ \hfill & =8\times {10}^{3+8}\hfill \\ \hfill & =8\times {10}^{11}\hfill \end{array}$
$\begin{array}{ll}(5\times {10}^{17})(8.1\times {10}^{-22})\hfill & =(5\times 8.1)({10}^{17}\times {10}^{-22})\hfill \\ \hfill & =40.5\times {10}^{17-22}\hfill \\ \hfill & =40.5\times {10}^{-5}\hfill \end{array}$
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.
$\begin{array}{l}40.5\times {10}^{-5}\hfill \\ 4.05\times {10}^{1}\times {10}^{-5}\hfill \\ 4.05\times ({10}^{1}\times {10}^{-5})\hfill \\ 4.05\times ({10}^{1-5})\hfill \\ 4.05\times {10}^{-4}\hfill \end{array}$
Thus,
$(5\times {10}^{17})(8.1\times {10}^{-22})=4.05\times {10}^{-4}$
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