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Very large and very small quantities and measures are often used to provide information in magazines, textbooks, television, newspapers and on the Internet. Some examples are:

  • The distance between the sun and Neptune is 4 500 000 000 km.
  • The diameter of an electron is approximately 0.000 000 000 000 22 inches.

Scientific notation is a convenient way to represent such numbers. How could you write the numbers above using scientific notation?

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Khan Academy Scientific Notation

Khan Academy Scientific Notation Examples

Guidance

To represent a number in scientific notation means to express the number as a product of two factors: a number between 1 and 10 (including 1) and a power of 10. A positive real number ‘ x ’ is said to be written in scientific notation if it is expressed as x=a \times 10^n where 1 \le a < 10 \ \text{and} \ n \ \varepsilon \ I. In other words, a number in scientific notation is a single nonzero digit followed by a decimal point and other digits, all multiplied by a power of 10.

When working with numbers written in scientific notation, you can use the following rules. These rules are proved by example in Example B and Example C.

\boxed{(A \times 10^n)+(B \times 10^n)=(A+B)\times 10^n}

\boxed{(A \times 10^n)-(B \times 10^n)=(A-B)\times 10^n}

\boxed{(A \times 10^m) \times (B \times 10^n)=(A \times B) \times (10^{m+n})}

\boxed{(A \times 10^m) \div (B \times 10^n)=(A \div B) \times (10^{m-n})}

Example A

Write the following numbers using scientific notation:

i) 2679000

ii) 0.00005728

Solutions:

i)

2679000&=2.679 \times 1000000\\2.679 \times 1000000&=2.679 \times 10^{{\color{red}6}}

The exponent, n = 6 , represents the decimal point that is 6 places to the right of the standard position of the decimal point.

ii)

0.00005728&=5.728 \times 0.00001\\5.728 \times 0.00001&=5.728 \times \frac{1}{100000}\\5.728 \times \frac{1}{100000}&=5.728 \times \frac{1}{10^{\color{red}5}}\\5.728 \times \frac{1}{100000}&=5.728 \times 10^{\color{red}-5}

The exponent, n = -5 , represents the decimal point that is 5 places to the left of the standard position of the decimal point.

One advantage of scientific notation is that calculations with large or small numbers can be done by applying the laws of exponents.

Example B

Complete the following table.

Expression in Scientific Notation Expression in Standard Form Result in Standard Form Result in Scientific Notation
1.3 \times 10^5+2.5 \times 10^5
3.7 \times 10^{-2}+5.1 \times 10^{-2}
4.6 \times 10^4-2.2 \times 10^4
7.9 \times 10^{-2}-5.4 \times 10^{-2}

Solution:

Expression in Scientific Notation Expression in Standard Form Result in Standard Form Result in Scientific Notation
1.3 \times 10^5+2.5 \times 10^5 130000+250000 380000 3.8 \times 10^5
3.7 \times 10^{-2}+5.1 \times 10^{-2} 0.037+0.051 0.088 8.8 \times 10^{-2}
4.6 \times 10^4-2.2 \times 10^4 46000-22000 24000 2.4 \times 10^4
7.9 \times 10^{-2}-5.4 \times 10^{-2} 0.079-0.054 0.025 2.5 \times 10^{-2}

Note that the numbers in the last column have the same power of 10 as those in the first column.

Example C

Complete the following table.

Expression in Scientific Notation Expression in Standard Form Result in Standard Form Result in Scientific Notation
(3.6 \times 10^2) \times (1.4 \times 10^3)
(2.5 \times 10^3) \times (1.1 \times 10^{-6})
(4.4 \times 10^4) \div (2.2 \times 10^2)
(6.8 \times 10^{-4}) \div (3.2 \times 10^{-2})

Solution:

Expression in Scientific Notation Expression in Standard Form Result in Standard Form Result in Scientific Notation
(3.6 \times 10^2) \times (1.4 \times 10^3) 360 \times 1400 504000 5.04 \times 10^5
(2.5 \times 10^3) \times (1.1 \times 10^{-6}) 2500 \times 0.0000011 0.00275 2.75 \times 10^{-3}
(4.4 \times 10^4) \div (2.2 \times 10^2) 44000 \div 220 200 2.0 \times 10^2
(6.8 \times 10^{-4}) \div (3.2 \times 10^{-2}) 0.00068 \div 0.032 0.02125 2.125 \times 10^{-2}

Note that for multiplication, the power of 10 is the result of adding the exponents of the powers in the first column. For division, the power of 10 is the result of subtracting the exponents of the powers in the first column.

Example D

Calculate each of the following:

i) 4.6 \times 10^4+5.3 \times 10^5

ii) 4.7 \times 10^{-3} - 2.4 \times 10^{-4}

iii) (7.3 \times 10^5) \times (6.8 \times 10^4)

iv) (4.8 \times 10^9) \div (5.79 \times 10^7)

Solution:

i) Before the rule \boxed{(A \times 10^n) + (B \times 10^n)=(A+B) \times 10^n} can be used, one of the numbers must be rewritten so that the powers of 10 are the same.

Rewrite 4.6 \times 10^4

4.6 \times 10^4=(0.46 \times 10^{\color{red}1}) \times 10^4 The power 10^{\color{red}1} indicates the number of places to the right that the decimal point must be moved to return 0.46 to the original number of 4.6.

(0.46 \times 10^1) \times 10^4=0.46 \times 10^{\color{red}5} Add the exponents of the power.

Rewrite the question and substitute 4.6 \times 10^4 with 0.46 \times 10^5 .

0.46 \times 10^5+5.3 \times 10^5

Apply the rule \boxed{(A \times 10^n)+(B \times 10^n)=(A+B) \times 10^n} .

&(0.46 \times 10^5) + (5.3 \times 10^5)=(0.46+5.3) \times 10^5\\&(0.46+5.3) \times 10^5=5.76 \times 10^5\\&\boxed{4.6 \times 10^4 + 5.3 \times 10^5=5.76 \times 10^5}

ii) 4.7 \times 10^{-3} -2.4 \times 10^{-4}

Before the rule \boxed{(A \times 10^n)-(B \times 10^n)=(A-B) \times 10^n} can be used, one of the numbers must be rewritten so that the powers of 10 are the same.

Rewrite 4.7 \times 10^{-3}

4.7 \times 10^{-3}=(4.7 \times 10^{{\color{red}-1}}) \times 10^{-3} The power 10^{{\color{red}-1}} indicates the number of places to the left that the decimal point must be moved to return 47 to the original number of 4.7.

(47 \times 10^{-1}) \times 10^{-3} = 47 \times 10^{{\color{red}-4}} Add the exponents of the power.

Rewrite the question and substitute 4.7 \times 10^{-3} with 47 \times 10^{-4} .

47 \times 10^{-4}-2.4 \times 10^{-4}

Apply the rule \boxed{(A \times 10^n)-(B \times 10^n)=(A-B) \times 10^n} .

(47 \times 10^{-4})-(2.4 \times 10^{-4})&=(47-2.4)\times 10^{-4}\\(47 \times 10^{-4})-(2.4 \times 10^{-4})&=44.6 \times 10^{-4}

The answer must be written in scientific notation.

&44.6 \times 10^{-4}=(4.46 \times 10^{{\color{red}1}}) \times 10^{-4} && \text{Apply the law of exponents }-\text{ add the  exponents of the power.}\\&4.46 \times 10 \times 10^{-4}=4.46 \times 10^{{\color{red}-3}}\\&\boxed{4.7 \times 10^{-3}-2.4 \times 10^{-4}=4.46 \times 10^{-3}}

iii) (7.3 \times 10^5) \times (6.8 \times 10^4)

&7.3 \times 10^5 \times 6.8 \times 10^4 && \text{Apply the rule } \boxed{(A \times 10^m) \times (B \times 10^n)=(A \times B) \times (10^{m+n})}.\\&(7.3 \times 10^5) \times (6.8 \times 10^4)=(7.3 \times 6.8) \times (10^{{\color{red}5+4}})\\&(7.3 \times 6.8) \times (10^{{\color{red}5+4}})=({\color{red}49.64}) \times (10^{{\color{red}9}})\\&({\color{red}49.64}) \times (10^{{\color{red}9}})=49.64 \times 10^9 && \text{Write the answer in scientific notation.}\\&49.64 \times 10^9=(4.964 \times 10^{{\color{red}1}}) \times 10^9 && \text{Apply the law of exponents }-\text{ add the exponents of the power.}\\&49.64 \times 10^9=4.964 \times 10^{{\color{red}10}}\\&\boxed{(7.3 \times 10^5) \times (6.8 \times 10^4)=4.964 \times 10^{10}}

iv) (4.8 \times 10^9) \div (5.79 \times 10^7)

&(4.8 \times 10^9) \div (5.79 \times 10^7) && \text{Apply the rule } \boxed{(A \times 10^m)\div(B \times 10^n)=(A \div B) \times (10^{m-n})}.\\&(4.8 \times 10^9) \div (5.79 \times 10^7) = (4.8 \div 5.79) \times 10^{{\color{red}9-7}} && \text{Apply the law of exponents }- \text{subtract the exponents of the power.}\\&(4.8 \div 5.79) \times 10^{{\color{red}9-7}}=(0.829) \times 10^{{\color{red}2}} && \text{Write the answer in scientific notation.}\\&(0.829) \times 10^{{\color{red}2}}=(8.29 \times 10^{{\color{red}-1}}) \times 10^{{\color{red}2}} && \text{Apply the law of exponents }-\text{ add the exponents of the power.}\\& \boxed{(8.29 \times 10^{{\color{red}-1}}) \times 10^2 = 8.29 \times 10^1}

Concept Problem Revisited

The distance between the sun and Neptune would be written as 4.5 \times 10^9 \ km and the diameter of an electron would be written as 2.2 \times 10^{-13} \ in .

Vocabulary

Scientific Notation
Scientific notation is a way of writing numbers in the form of a number between 1 and 10 multiplied by a power of 10. The number 196.5 written in scientific notation is 1.965 \times 10^2 and the number 0.0760 written in scientific notation is 7.60 \times 10^{-2} .

Guided Practice

1. Express the following product in scientific notation: (4 \times 10^{12})(9.2 \times 10^7)

2. Express the following quotient in scientific notation: \frac{6400000}{0.008}

3. If a&=0.000415\\b&=521\\c&=71640

find an approximate value for \frac{ab}{c} . Express the answer in scientific notation.

Answers:

1. (4 \times 10^{12})(9.2 \times 10^7)

Apply the rule \boxed{(A \times 10^m) \times (B \times 10^n) =(A \times B) \times (10^{m+n})}

&(4 \times 10^{12}) \times (9.2 \times 10^7) = (4 \times 9.2) \times (10^{12+7})\\&(4 \times 9.2) \times (10^{{\color{red}12+7}})={\color{red}36.8} \times 10^{{\color{red}19}}

Express the answer in scientific notation.

& 36.8 \times 10^{19}=(3.68 \times 10^{{\color{red}1}}) \times 10^{{\color{red}19}}\\&(3.68 \times 10^{{\color{red}1}}) \times 10^{{\color{red}19}}=3.68 \times 10^{{\color{red}20}}\\& \boxed{(4 \times 10^{12})(9.2 \times 10^7)=3.68 \times 10^{20}}

2. \frac{6400000}{0.008}

Begin by expressing the numerator and the denominator in scientific notation.

\frac{6.4 \times 10^6}{8.0 \times 10^{-3}}

Apply the rule \boxed{(A \times 10^m) \div (B \times 10^n)=(A \div B) \times (10^{m+n})} .

&(6.4 \times 10^6) \div (8.0 \times 10^{-3})=({\color{red}6.4 \div 8.0}) \times (10^{{\color{red}6--3}}) && \text{Apply the law of exponents }-\text{ subtract the exponents of the powers.}\\&(6.4 \div 8.0) \times (10^{6--3})=({\color{red}0.8}) \times (10^{{\color{red}9}})\\&({\color{red}0.8}) \times (10^{\color{red}9})=0.8 \times 10^9 && \text{Express the answer in scientific notation.}\\&0.8 \times 10^9=({\color{red}8.0} \times 10^{\color{red}-1}) \times 10^9\\&0.8 \times 10^9=8.0 \times 10^{-1} \times 10^9 && \text{Apply the law of exponents }-\text{ add the exponents of the powers.}\\& 8.0 \times 10^{-1} \times 10^9 =8.0 \times 10^8\\& \boxed{\frac{6400000}{0.008}=8.0 \times 10^8} && \text{Express the answer in scientific notation.}

3.

a&=0.000415\\b&=521\\c&=71640

Express all values in scientific notation.

0.000415 &= 4.15 \times 10^{-4}\\521 &= 5.21 \times 10^2\\71640 &= 7.1640 \times 10^4

Use the values in scientific notation to determine an approximate value for \frac{ab}{c} .

\frac{ab}{c}=\frac{(4.15 \times 10^{-4})(5.21 \times 10^2)}{7.1640 \times 10^4}

In the numerator, apply the rule \boxed{(A \times 10^m) \times (B \times 10^n)=(A \times B) \times (10^{m+n})}

&\frac{(4.15 \times 10^{-4})(5.21 \times 10^2)}{7.1640 \times 10^4}=\frac{(4.15 \times 5.21) \times (10^{-4} \times 10^2)}{7.1640 \times 10^4}\\&\frac{(4.15 \times 5.21) \times (10^{-4} \times 10^2)}{7.1640 \times 10^4}=\frac{21.6215 \times 10^{-2}}{7.1640 \times 10^4}\\& \text{Apply the rule} \ \boxed{(A \times 10^m) \div (B \times 10^n)=(A \div B) \times (10^{m-n})}.\\& \frac{21.6215 \times 10^{-2}}{7.1640 \times 10^4}=(21.6215 \div 7.1640) \times (10^{-2} \times 10^4)\\& \boxed{(21.6215 \div 7.1640) \times (10^{-2} \times 10^4)=3.018 \times 10^{-6}}

Practice

Express each of the following in scientific notation:

  1. 42000
  2. 0.00087
  3. 150.64
  4. 56789
  5. 0.00947

Express each of the following in standard form:

  1. 4.26 \times 10^5
  2. 8 \times 10^4
  3. 5.967 \times 10^{10}
  4. 1.482 \times 10^{-6}
  5. 7.64 \times 10^{-3}

Perform the indicated operations and express the answer in scientific notation

  1. 8.9 \times 10^4+4.3 \times 10^5
  2. 8.7 \times 10^{-4} -6.5 \times 10^{-5}
  3. (5.3 \times 10^6) \times (7.9 \times 10^5)
  4. (3.9 \times 10^8) \div (2.8 \times 10^6)

For the given values, perform the indicated operations for \frac{ab}{c} and express the answer in scientific notation and standard form.

a&=76.1\\b&=818000000\\c&=0.000016

a &=9.13 \times 10^9\\b &=5.45 \times 10^{-23}\\c &=1.62

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Jan 29, 2013

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