Example2.5.1
Simplify \((4 \times 10^{-7})(2 \times 10^9)\)āwrite the result using both scientific notation and standard notation.
Numbers written in scientific notation have the form \(a \times 10^n\) where \(1 \leq \abs a \lt 10\text{.}\) The value of \(m\) is always an integer. Two simple examples of numbers written using both standard notation and scientific notation are shown below
The basic concept behind conversion between standard notation and scientific notation is that every action requires an equal but opposite action. To wit:
Moving the decimal point left one digit corresponds to the power of \(10\) increasing by one.
Moving the decimal point right one digit corresponds to the power of \(10\) decreasing by one.
Suppose that we were charged with writing \(160,000\) using scientific notation. Because the number before the multiplication sign needs to be between \(1\) and \(10\text{,}\) we need to move the decimal point from the position found in \(160000.\) to the position found in \(1.60000\text{,}\) which is a leftward shift of five digits. This means that the power of \(10\) needs to be increased by five.
Rewriting \(0.000762\) requires shifting the decimal point to the right four digits, which corresponds to a decrease of four in the power of ten.
Now suppose that our objective was to write \(-3.8 \times 10^{-7}\) in standard notation. In order for the number to be written in standard notation, the power of \(10\) needs to \(0\text{.}\) For example,
So when writing \(3.8 \times 10^{-7}\) in standard notation, we need to increase the power of \(10\) by \(7\) which corresponds to a leftward shift of the decimal point by \(7\) digits. To achieve this end, we introduce six zeros to the left of the \(3\text{.}\) That is:
When multiplying or dividing numbers written in scientific notation, there are three tasks to complete.
Multiply or divide the numbers that appear before the scientific notation multiplication signs. If the absolute value of the result in not between \(1\) and \(10\text{,}\) make sure that you don't skip step \(3\text{!}\)
Add or subtract the powers of \(10\text{,}\) dependent upon whether you are multiplying or dividing the numbers.
If necessary, adjust the decimal point and power of \(10\) so that the final result has a number before the multiplication sign whose absolute value falls between \(1\) and \(10\text{.}\) If you have to choose between \(1\) and \(10\text{,}\) always go with \(1\text{.}\)
If you are in a science class, you also need to consider significant digits during this process.
Let's see a few of examples.
Simplify \((4 \times 10^{-7})(2 \times 10^9)\)āwrite the result using both scientific notation and standard notation.
Simplify \(\frac{7.0 \times 10^7}{5.0 \times 10^{-2}}\)āwrite the result using both scientific notation and standard notation.
Simplify \(\frac{4 \times 10^3}{8 \times 10^{10}}\)āwrite the result using both scientific notation and standard notation.
Note that we aren't quite finished. The number before the multiplication sign needs to be between \(1\) and \(10\text{,}\) so we need to move the decimal point one digit to the right. To balance that act, we need to decrease the exponent on \(10\) by one. Putting it all together we have:
Write each number in scientific notation.
Write each number in standard notation.
Determine each product or quotientāwrite the results using both scientific notation and standard notation.
\((2.0\times10^{-7})(3.0\times10^{12})\)
\(\begin{aligned}[t] (2.0\times10^{-7})(3.0\times10^{12})\amp=(2.0\times 3.0)\times(10^{-7}\times10^{12})\\ \amp=6.0\times10^5\\ \amp=600,000 \end{aligned}\)
\((4.0\times10^{-1})(3.0\times10^{-9})\)
\(\begin{aligned}[t] (4.0\times10^{-1})(3.0\times10^{-9})\amp=(4.0\times 3.0)\times({10^{-1}\times10^{-9}})\\ \amp=12\times10^{-10}\\ \amp=1.2\times10^{-9}\\ \amp=0.0000000012 \end{aligned}\)
\(\frac{1.2\times10^7}{2.4\times10^{-3}}\)
\(\begin{aligned}[t] \frac{1.2\times10^7}{2.4\times10^{-3}}\amp=\frac{1.2}{2.4}\times\frac{10^7}{10^{-3}}\\ \amp=0.50\times10^{10}\\ \amp=5.0\times10^9\\ \amp=5,000,000,000 \end{aligned}\)
\(\frac{9.0\times10^{-5}}{2.0\times10^{-2}}\)
\(\begin{aligned}[t] \frac{9.0\times10^{-5}}{2.0\times10^{-2}}\amp=\frac{9.0}{2.0}\times\frac{10^{-5}}{10^{-2}}\\ \amp=4.5\times10^{-3}\\ \amp=0.0045 \end{aligned}\)
\((8.0\times10^7)(8.0\times10^{-7})\)
\(\begin{aligned}[t] (8.0\times10^7)(8.0\times10^{-7})\amp=(8.0\times 8.0)\times(10^7\times10^{-7})\\ \amp=64\times10^0\\ \amp=6.4\times10^1\\ \amp=64 \end{aligned}\)
\(\frac{3.0\times{10^5}}{9.0\times10^{-9}}\)
\(\begin{aligned}[t] \frac{3.0\times10^5}{9.0\times10^{-9}}\amp=\frac{3.0}{9.0}\times\frac{10^5}{10^{-9}}\\ \amp=0.33\times10^{14}\\ \amp=3.3\times10^{13}\\ \amp=33,000,000,000,000 \end{aligned}\)
An application problem.
The speed of light is approximately \(3.00\times10^8\) m/s and there are approximately \(1.61\times10^3\) meters in one mile. Use this information to answer each of the following questions.
(\(3.00\times10^8\) m/s)(\(1\) hr)(\(60\) min/hr)(\(60\) s/min)\(=1.08\times10^{12}\) m
Light travels approximately \(1.08\times10^{12}\) meters in one hour.
(\(1.08\times10^{12}\) m)\(\left(\frac{1}{1.61\times10^3}\,\text{mi/m}\right)=6.71\times10^8\) mi.
Light travels approximately \(6.71\times10^8\) miles in one hour. For reference, that's a little less than the average distance between Earth and Saturn which in turn is roughly 8.5 times the average distance between Earth and the sun.