7.2: Scientific Notation
( \newcommand{\kernel}{\mathrm{null}\,}\)
We begin this section by examining powers of ten.
101=10102=10⋅10=100103=10⋅10⋅10=1,000104=10⋅10⋅10⋅10=10,000
Note that the answer for 103 is a one followed by three zeros. The answer for 104 is a one followed by four zeros. Do you see the pattern?
Nonnegative powers of ten
In the expression 10n, the exponent matches the number of zeros in the answer. Hence, 10n will be a 1 followed by n zeros.
Example 7.2.1
Simplify: 109.
Solution
109 should be a 1 followed by 9 zeros.109=1,000,000,000
Exercise 7.2.1
Simplify: 106.
- Answer
-
1,000,000
Next, let’s examine negative powers of ten.
10−1=110=0.110−2=1100=0.0110−3=11000=0.00110−4=110000=0.0001
Note that the answer for 10−3 has three decimal places and the answer for 10−4 contains four decimal places.
Negative powers of ten
In the expression 10−n, the exponent matches the number of decimal places in the answer. Hence, 10−n will have n decimal places, the first n−1 of which are zeros and the digit in the nth decimal place is a 1.
Example 7.2.2
Simplify: 10−7.
Solution
10−7 should have seven decimal places, the first six of which are zeros, and the digit in the seventh decimal place is a 1.10−7=0.0000001
Exercise 7.2.2
Simplify: 10−5.
- Answer
-
0.00001
Multiplying Decimal Numbers by Powers of Ten
Let’s multiply 1.234567 by 103, or equivalently, by 1,000.
1.234567×10001234.567000
The sum total of digits to the right of the decimal point in 1.234567 and 1000 is 6. Therefore, we place the decimal point in the product so that there are six digits to the right of the decimal point.
However, the trailing zeros may be removed without changing the value of the product. That is, 1.234567 times 1000 is 1234.567. Note that the decimal point in the product is three places further to the right than in the original factor. This observation leads to the following result.
Multiplying by a nonnegative power of ten
Multiplying a decimal number by 10n, where n=0,1,2,3,…, will move the decimal point n places to the right.
Example 7.2.3
Simplify: 325.6783×102.
Solution
Multiplying by 102 will move the decimal point two places to the right. Thus:325.6783×102=32,567.83
Exercise 7.2.3
Simplify: 23.57889×103
- Answer
-
23,578.89
Example 7.2.4
Simplify: 1.25×105.
Solution
Multiplying by 105 will move the decimal point two places to the right. In this case, we need to add zeros at the end of the number to accomplish moving the decimal 5 places to the right.1.25×105=125,000
Exercise 7.2.14
Simplify: 2.35×104
- Answer
-
23,500
Let’s multiply 453.9 by 10−2, or equivalently, by 0.01.
453.9×0.014.539
The sum total of digits to the right of the decimal point in 453.9 and 0.01 is 3. Therefore, we place the decimal point in the product so that there are 3 digits to the right of the decimal point. That is, 453.9×10−2=4.539. Note that the decimal point in the product is two places further to the left than in the original factor. This observation leads to the following result.
Multiplying by a negative power of ten
Multiplying a decimal number by 10−n, where n=1,2,3,…, will move the decimal point n places to the left.
Example 7.2.5
Simplify: 14,567.8×10−3.
Solution
Multiplying by 10−3 will move the decimal point three places to the left. Thus: 14,567.8×10−3=14.5678
Exercise 7.2.5
Simplify: 3,854.2×10−1
- Answer
-
385.42
Example 7.2.6
Simplify: 4.3×10−4.
Solution
Multiplying by 10−4 will move the decimal point four places to the left. In this case, we need to add some leading zeros at the beginning of the number to accomplish moving the decimal 4 places to the left.4.3×10−4=0.00043. Note also the leading zero before the decimal point. Although .00043 is an equivalent number, the form 0.00043 is preferred in mathematics and science.
Exercise 7.2.6
Simplify: 2.2×10−2
- Answer
-
0.022
Scientific Notation Form
We start by defining the form of a number called scientific notation.
Scientific notation
A number having the forma×10bwhere b is an integer and 1≤|a|<10, is said to be in scientific notation.
The requirement 1≤|a|<10 says that the magnitude of a must be a least 1 and less than 10.
- The number 12.34×10−4 is not in scientific notation because |12.34|=12.34 is larger than 10.
- The number −0.95×103 is not in scientific notation because |−0.95|=0.95 is less than 1.
- The number 7.58×10−12 is in scientific notation because |7.58|=7.58 is greater than or equal to 1 and less than 10.
- The number −1.0×1015 is in scientific notation because |−1.0|=1.0 is greater than or equal to 1 and less than 10.
After contemplating these examples, it follows that a number in scientific notation should have exactly one of the digits 1,2,3,…,9 before the decimal point. Exactly one, no more, no less. Thus, each of the following numbers is in scientific notation.
4.7×108,−3.764×10−1,3.2×100,and−1.25×10−22
Placing a Number in Scientific Notation
To place a number into scientific notation, we need to move the decimal point so that exactly one of the digits 1,2,3,…,9 remains to the left of the decimal point, then multiply by the appropriate power of 10 so that the result is equivalent to the original number.
Example 7.2.7
Place the number 1,234 in scientific notation.
Solution
Move the decimal point three places to the left so that it is positioned just after the 1. To make this new number equal to 1,234, multiply by 103. Thus:1,234=1.234×103
Check: Multiplying by 103 moves the decimal three places to the right, so:1.234×103=1,234 This is the original number, so our scientific notation form is correct.
Exercise 7.2.7
Place the number 54,321 in scientific notation.
- Answer
-
5.4321×104
Example 7.2.8
Place the number 0.000025 in scientific notation.
Solution
Move the decimal point five places to the right so that it is positioned just after the 2. To make this new number equal to 0.000025, multiply by 10−5. Thus: 0.000025=2.5×10−5
Check: Multiplying by 10−5 moves the decimal five places to the left, so:2.5×10−5=0.000025This is the original number, so our scientific notation form is correct.
Exercise 7.2.8
Place the number 0.0175 in scientific notation.
- Answer
-
1.75×10−2
Example 7.2.9
Place the number 34.5×10−11 in scientific notation.
Solution
First, move the decimal point one place to the left so that it is positioned just after the three. To make this new form equal to 34.5, multiply by 101.34.5×10−11=3.45×101×10−11Now, repeat the base 10 and add the exponents.
=3.45×10−10
Exercise 7.2.9
Place the number 756.98×10−5 in scientific notation.
- Answer
-
7.5698×10−3
Example 7.2.10
Place the number 0.00093×1012 in scientific notation.
Solution
First, move the decimal point four places to the right so that it is positioned just after the nine. To make this new form equal to 0.00093, multiply by 10−4. 0.00093×1012=9.3×10−4×1012
Now, repeat the base 10 and add the exponents.=9.3×108
Exercise 7.2.10
Place the number 0.00824×108 in scientific notation.
- Answer
-
8.24×105
Scientific Notation and the Graphing Calculator
The TI-84 graphing calculator has a special button for entering numbers in scientific notation. Locate the “comma” key just about the number 7 key on the calculator’s keyboard (see Figure 7.2.1). Just above the “comma” key, printed on the calculator’s case is the symbol EE. It’s in the same color as the 2nd key, so you’ll have to use the 2nd key to access this symbol.

We know that 2.3×102=230. Let’s see if the calculator gives the same interpretation.
- Enter 2.3.
- Press the 2nd key, then the comma key. This will put E on the calculator view screen.
- Enter a 2.
- Press ENTER.

The result of these steps is shown in the first image in Figure 7.2.2. Note that the calculator interprets 2.3E2 as 2.3×102 and gives the correct answer, 230. You can continue entering numbers in scientific notation (see the middle image in Figure 7.2.2). However, at some point the numbers become too large and the calculator responds by outputting the numbers in scientific notaiton. You can also force your calculator to display numbers in scientific notation in all situations, by first pressing the MODE key, then selecting SCI on the first line and pressing the ENTER key (see the third image in Figure 7.2.2). You can return your calculator to “normal” mode by selecting NORMAL and pressing the ENTER key.
Example 7.2.11
Use the graphing calculator to simplify: (2.35×10−12)(3.25×10−4)
Solution
First, note that we can approximate (2.35×10−12)(3.25×10−4) by taking the product of 2 and 3 and adding the powers of ten.
(2.35×10−12)(3.25×10−4)≈(2×10−12)(3×10−4)Approximate: 2.35≈2 and 3.25≈3≈6×10−162⋅3=6 and 10−12⋅10−4=10−16
The graphing calculator will provide an accurate answer. Enter 2.35E−12, press the “times” button, then enter 3.25E-4 and press the ENTER button. Be sure to use the “negate” button and not the “subtract” button to produce the minus sign. The result is shown in Figure 7.2.3.

Thus, (2.35×10−12)(3.25×10−4)=7.6375×10−16. Note that this is fairly close to our estimate of 6×10−16.
Exercise 7.2.11
Use the graphing calculator to simplify: (3.42×106)(5.86×10−9)
- Answer
-
2.00412×10−2
Reporting your answer on your homework
After computing the answer to Example 7.2.11 on your calculator, write the following on your homework:(2.35×10−12)(3.25×10−4)=7.6375×10−16Do not write 7.6375E−16.
Example 7.2.12
Use the graphing calculator to simplify:6.1×10−3(2.7×104)(1.1×108)
Solution
Again, it is not difficult to produce an approximate answer.
6.1×10−3(2.7×104)(1.1×108)≈6×10−3(3×104)(1×108)6.1≈6,2.7≈3, and 1.1≈1≈6×10−33×10123⋅1=3 and 104⋅108=1012≈63⋅10−31012acbd=ab⋅cd≈2×10−1563=2 and 10−31012=10−15
Let’s get a precise answer with our calculator. Enter the numerator as 6.1E3, then press the “division” button. Remember that we must surround the denominator with parentheses. So press the open parentheses key, then enter 2.7E4. Press the “times” key, then enter 1.1E8. Press the close parentheses key and press the ENTER button. The result is shown in Figure 7.2.4.

Thus, 6.1×10−3/(2.7×104×1.1×108)=2.05387205×10−15. Note that this is fairly close to our estimate of 2×10−15.
Exercise 7.2.12
Use the graphing calculator to simplify: 2.6×104(7.1×10−2)(6.3×107)
- Answer
-
5.8126537×10−3
Example 7.2.13
Isaac Newton’s universal law of gravitation is defined by the formula F=GmMr2 where F is the force of attraction between two objects having mass m and M, r is the distance between the two objects, and G is Newton’s gravitational constant defined by:G=6.67428×10−11N(m/kg)2Given that the mass of the moon is 7.3477×1022 kilograms (kg), the mass of the earth is 5.9736×1024 kilograms (kg), and the average distance between the moon and the earth is 3.84403×108 meters (m), find the force of attraction between the earth and the moon (in newtons (N)).
Solution
Plug the given numbers into Newton’s universal law of gravitation.
F=GmMr2
F=(6.673×10−11)(7.3477×1022)(5.9736×1024)(3.84403×108)2
Enter the expression into your calculator (see Figure 7.2.5) as:
(6.673E−11∗7.3477E22∗5.9736E24)/(3.84403E8)∧2

Hence, the force of attraction between the earth and the moon is approximately 1.98×1020 newtons (N).
Exercise 7.2.13
The mass of the International Space Station is 450,000 kg, and its average distance to the center of the earth is 387,000 m. Find the force of attraction between the earth and the station (in newtons (N)).
- Answer
-
≈1.20×109N
Example 7.2.14
The closest star to the earth is Alpha Centauri, 4.37 light-years from the earth. A light-year is the distance that light will travel in one-year’s time. The speed of light is 186,000 miles per second. How many miles from the earth is Alpha Centauri?
Solution
Because the speed of light is measured in miles per second, let’s first compute the number of seconds in 4.37 years. Because there are 365 days in a year, 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute, we can write:
4.37yr=4.37yr×365dayyr×24hrday×60minhr×60smin=4.37⧸yr×365⧸day⧸yr×24⧸hr⧸day×60⧸min⧸hr×60s⧸min
Note how the units cancel, indicating that the final answer is in seconds. With our calculator mode set to scientific notation (see the image on the right in Figure 7.2.2), we multiply the numbers to get the result shown in Figure 7.2.6. Rounding, the number of seconds in 4.37 years is approximately 1.38×108 seconds.
Next, we compute the distance the light travels in 4.37 years. Using the fact that the distance traveled equals the speed times the time traveled, we have:
Distance=Speed×Time=1.86×105mis⋅1.38×108s=1.86×105mi⧸s⋅1.38×108⧸s
Note how the units cancel, indicating that our answer is in miles. Again, with our calculator set in scientific notation mode, we compute the product of 1.86×105 and 1.38×108. The result is shown in the image on the right in Figure 7.2.6.

Thus, the star Alpha Centauri is approximately 2.5668×1013 miles from the earth, or 2.5668×1013miles≈25,668,000,000,000milespronounced “twenty-five quadrillion, six hundred sixty-eight trillion miles.”
Exercise 7.2.14
The star Sirius is 8.58 light-years from the earth. How many miles from the earth is Sirius?
- Answer
-
≈5.2425×1013 miles