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Mathematics LibreTexts

7.2: Scientific Notation

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We begin this section by examining powers of ten.

101=10102=1010=100103=101010=1,000104=10101010=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.

101=110=0.1102=1100=0.01103=11000=0.001104=110000=0.0001

Note that the answer for 103 has three decimal places and the answer for 104 contains four decimal places.

Negative powers of ten

In the expression 10n, the exponent matches the number of decimal places in the answer. Hence, 10n will have n decimal places, the first n1 of which are zeros and the digit in the nth decimal place is a 1.

Example 7.2.2

Simplify: 107.

Solution

107 should have seven decimal places, the first six of which are zeros, and the digit in the seventh decimal place is a 1.107=0.0000001

Exercise 7.2.2

Simplify: 105.

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 102, 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, 453.9×102=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 10n, where n=1,2,3,, will move the decimal point n places to the left.

Example 7.2.5

Simplify: 14,567.8×103.

Solution

Multiplying by 103 will move the decimal point three places to the left. Thus: 14,567.8×103=14.5678

Exercise 7.2.5

Simplify: 3,854.2×101

Answer

385.42

Example 7.2.6

Simplify: 4.3×104.

Solution

Multiplying by 104 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.4.3×104=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: 2.2×102

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 12.34×104 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 7.58×1012 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.

4.7×108,3.764×101,3.2×100,and1.25×1022

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 105. Thus: 0.000025=2.5×105

Check: Multiplying by 105 moves the decimal five places to the left, so:2.5×105=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

1.75×102

Example 7.2.9

Place the number 34.5×1011 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 101.34.5×1011=3.45×101×1011Now, repeat the base 10 and add the exponents.

=3.45×1010

Exercise 7.2.9

Place the number 756.98×105 in scientific notation.

Answer

7.5698×103

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 104. 0.00093×1012=9.3×104×1012

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.

fig 7.2.1.png
Figure 7.2.1: Locate the EE label above the “comma” key on the keyboard.

We know that 2.3×102=230. Let’s see if the calculator gives the same interpretation.

  1. Enter 2.3.
  2. Press the 2nd key, then the comma key. This will put E on the calculator view screen.
  3. Enter a 2.
  4. Press ENTER.
fig 7.2.2.png
Figure 7.2.2: Entering numbers in scientific notation.

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: (2.35×1012)(3.25×104)

Solution

First, note that we can approximate (2.35×1012)(3.25×104) by taking the product of 2 and 3 and adding the powers of ten.

(2.35×1012)(3.25×104)(2×1012)(3×104)Approximate: 2.352 and 3.2536×101623=6 and 1012104=1016

The graphing calculator will provide an accurate answer. Enter 2.35E12, 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.

fig 7.2.3.png
Figure 7.2.3: Computing (2.35×1012)(3.25×104).

Thus, (2.35×1012)(3.25×104)=7.6375×1016. Note that this is fairly close to our estimate of 6×1016.

Exercise 7.2.11

Use the graphing calculator to simplify: (3.42×106)(5.86×109)

Answer

2.00412×102

Reporting your answer on your homework

After computing the answer to Example 7.2.11 on your calculator, write the following on your homework:(2.35×1012)(3.25×104)=7.6375×1016Do not write 7.6375E16.

Example 7.2.12

Use the graphing calculator to simplify:6.1×103(2.7×104)(1.1×108)

Solution

Again, it is not difficult to produce an approximate answer.

6.1×103(2.7×104)(1.1×108)6×103(3×104)(1×108)6.16,2.73, and 1.116×1033×101231=3 and 104108=1012631031012acbd=abcd2×101563=2 and 1031012=1015

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.

fig 7.2.4.png
Figure 7.2.4: Computing 6.1×103/(2.7×104×1.1×108).

Thus, 6.1×103/(2.7×104×1.1×108)=2.05387205×1015. Note that this is fairly close to our estimate of 2×1015.

Exercise 7.2.12

Use the graphing calculator to simplify: 2.6×104(7.1×102)(6.3×107)

Answer

5.8126537×103

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=6.67428×1011N(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=(6.673×1011)(7.3477×1022)(5.9736×1024)(3.84403×108)2

Enter the expression into your calculator (see Figure 7.2.5) as:

(6.673E117.3477E225.9736E24)/(3.84403E8)2

fig 7.2.5.png
Figure 7.2.5: Computing force of attraction between earth and the moon.

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=4.37yr×365dayyr×24hrday×60minhr×60smin

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=1.86×105mis1.38×108s=1.86×105mis1.38×108s

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.

fig 7.2.6.png
Figure 7.2.6: Computing the distance to Alpha Centauri in miles.

Thus, the star Alpha Centauri is approximately 2.5668×1013 miles from the earth, or 2.5668×1013miles25,668,000,000,000milespronounced “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


This page titled 7.2: Scientific Notation is shared under a CC BY-NC-ND 3.0 license and was authored, remixed, and/or curated by David Arnold via source content that was edited to the style and standards of the LibreTexts platform.

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