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Section 3.5 Scientific 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. The number \(a\) is called the coefficient. Two simple examples of numbers written using both standard notation and scientific notation are shown below.

\begin{equation*} -300=-3 \times 10^2 \end{equation*}
\begin{equation*} 0.45=4.5 \times 10^{-1} \end{equation*}

You can use Figure 3.5.1 to explore the effect of changing the exponent on 10. Make sure that you keep track of the number of digits you have to cross to get back to 3.2. You can use Figure 3.5.2 to practice rewriting numbers using scientific notation.

Figure 3.5.1. Use the slider to change the exponents on 10.
Figure 3.5.2. Rewrite the given number using scientific notation.

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.

\begin{equation*} 236.7 \times 10^0=23.67 \times 10^1=2.367 \times 10^2 \end{equation*}

Moving the decimal point right one digit corresponds to the power of \(10\) decreasing by one.

\begin{equation*} 0.056 \times 10^0=0.56 \times 10^{-1}=5.6 \times 10^{-2} \end{equation*}

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.

\begin{align*} 160,000\amp=160,000 \times 10^0\\ \amp=1.6 \times 10^5 \end{align*}

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.

\begin{align*} 0.000762\amp=0.000762 \times 10^0\\ \amp=7.62 \times 10^{-4} \end{align*}

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,

\begin{align*} 7 \times 10^0\amp=7 \times 1\\ \amp=7 \end{align*}

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:

\begin{equation*} 3.8 \times 10^{-7}=.00000038 \end{equation*}

When multiplying or dividing numbers written in scientific notation, there are three tasks to complete.

  1. 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 \(3\text{!}\)

  2. Add or subtract the powers of \(10\text{,}\) dependent upon whether you are multiplying or dividing the numbers.

  3. 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.

Example 3.5.3.

Simplify \((4 \times 10^{-7})(2 \times 10^9)\)—write the result using both scientific notation and standard notation.

Solution
\begin{align*} (4 \times 10^{-7})(2 \times 10^9)\amp=(4 \times 2) \times (10^{-7} \times 10^9)\\ \amp=8 \times 10^2 \end{align*}
Example 3.5.4.

Simplify \(\frac{7.0 \times 10^7}{5.0 \times 10^{-2}}\)—write the result using both scientific notation and standard notation.

Solution
\begin{align*} \frac{7.0 \times 10^7}{5.0 \times 10^{-2}}\amp=\frac{7.0}{5.0} \times \frac{10^7}{10^{-2}}\\ \amp=1.4 \times 10^9 \end{align*}
Example 3.5.5.

Simplify \(\frac{4 \times 10^3}{8 \times 10^{10}}\)—write the result using both scientific notation and standard notation.

Solution
\begin{align*} \frac{4 \times 10^3}{8 \times 10^{10}}\amp=\left(\frac{4}{8}\right) \times \frac{10^3}{10^{10}}\\ \amp=0.5 \times 10^{-7} \end{align*}

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:

\begin{align*} \frac{4 \times 10^3}{8 \times 10^{10}}\amp=\left(\frac{4}{8}\right) \times \frac{10^3}{10^{10}}\\ \amp=0.5 \times 10^{-7}\\ \amp=5 \times 10^{-8} \end{align*}

You can use Figure 3.5.6 to practice multiplying numbers stated in scientific notation. Make sure that the coefficient in our answer lies between one and ten.

Figure 3.5.6. Positive integer exponents keep track of the number of factors in the expression.

Exercises Exercises

Write each number in scientific notation.

1.

\(42,000\)

Solution

\(42,000=4.2\times 10^4\)

2.

\(-0.000000714\)

Solution

\(-0.000000714=-7.14\times 10^{-7}\)

3.

\(0.00067\)

Solution

\(0.00067=6.7\times 10^{-4}\)

4.

\(-357,288\)

Solution

\(-357,288=-3.57288\times 10^5\)

Write each number in standard notation.

5.

\(-8.6\times 10^{-7}\)

Solution

\(-8.6\times 10^{-7}=-0.00000086\)

6.

\(-3.7\times 10^9\)

Solution

\(-3.7\times 10^9-3,700,000,000\)

7.

\(4.55\times 10^{-3}\)

Solution

\(4.55\times 10^{-3}=0.00455\)

8.

\(2.7\times 10^5\)

Solution

\(2.7\times 10^5=270,000\)

Determine each product or quotient—write the results using both scientific notation and standard notation.

9.

\((2.0\times10^{-7})(3.0\times10^{12})\)

Solution

\(\begin{aligned}[t] (2.0\times10^{-7})(3.0\times10^{12})\amp=(2.0\times 3.0)\times(10^{-7}\times10^{12})\\ \amp=6.0\times10^5\\ \amp=600,000 \end{aligned}\)

10.

\((4.0\times10^{-1})(3.0\times10^{-9})\)

Solution

\(\begin{aligned}[t] (4.0\times10^{-1})(3.0\times10^{-9})\amp=(4.0\times 3.0)\times({10^{-1}\times10^{-9}})\\ \amp=12\times10^{-10}\\ \amp=1.2\times10^{-9}\\ \amp=0.0000000012 \end{aligned}\)

11.

\(\frac{1.2\times10^7}{2.4\times10^{-3}}\)

Solution

\(\begin{aligned}[t] \frac{1.2\times10^7}{2.4\times10^{-3}}\amp=\frac{1.2}{2.4}\times\frac{10^7}{10^{-3}}\\ \amp=0.50\times10^{10}\\ \amp=5.0\times10^9\\ \amp=5,000,000,000 \end{aligned}\)

12.

\(\frac{9.0\times10^{-5}}{2.0\times10^{-2}}\)

Solution

\(\begin{aligned}[t] \frac{9.0\times10^{-5}}{2.0\times10^{-2}}\amp=\frac{9.0}{2.0}\times\frac{10^{-5}}{10^{-2}}\\ \amp=4.5\times10^{-3}\\ \amp=0.0045 \end{aligned}\)

13.

\((8.0\times10^7)(8.0\times10^{-7})\)

Solution

\(\begin{aligned}[t] (8.0\times10^7)(8.0\times10^{-7})\amp=(8.0\times 8.0)\times(10^7\times10^{-7})\\ \amp=64\times10^0\\ \amp=6.4\times10^1\\ \amp=64 \end{aligned}\)

14.

\(\frac{3.0\times{10^5}}{9.0\times10^{-9}}\)

Solution

\(\begin{aligned}[t] \frac{3.0\times10^5}{9.0\times10^{-9}}\amp=\frac{3.0}{9.0}\times\frac{10^5}{10^{-9}}\\ \amp=0.33\times10^{14}\\ \amp=3.3\times10^{13}\\ \amp=33,000,000,000,000 \end{aligned}\)

An application problem.

15.

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.

  1. Determine (to three significant digits) the number of meters that light travels in one hour.
  2. Determine (to three significant digits) the number of miles that light travels in one hour.
Solution
  1. (\(3.00\times10^8\) 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.

  2. (\(1.08\times10^{12}\) 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.