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Zero, Negative, and Fractional Exponents

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Negative and Zero Exponents

The magnitude of an earthquake represents the exponent m in the expression 10^m .

Valdivia, Chile has suffered two major earthquakes. The 1575 Valdivia earthquake had a magnitude of 8.5. The world's largest earthquake was the 1960 Valdivia earthquake at a magnitude of 9.5.

What was the size of the 1575 earthquake compared to the 1960 one?

Source: http://en.wikipedia.org/wiki/List_of_earthquakes

Guidance

In this concept, we will introduce negative and zero exponents. First, let’s address a zero in the exponent through an investigation.

Investigation: Zero Exponents

1. Evaluate \frac{5^6}{5^6} by using the Quotient of Powers property.

\frac{5^6}{5^6} = 5^{6-6} = 5^0

2. What is a number divided by itself? Apply this to #1.

\frac{5^6}{5^6} = 1

3. Fill in the blanks. \frac{a^m}{a^m} = a^{m-m} = a^- = _-

a^0 = 1

Investigation: Negative Exponents

1. Expand \frac{3^2}{3^7} and cancel out the common 3’s and write your answer with positive exponents.

\frac{3^2}{3^7} = \frac{\cancel{3} \cdot \cancel{3}}{\cancel{3} \cdot \cancel{3} \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3} = \frac{1}{3^5}

2. Evaluate \frac{3^2}{3^7} by using the Quotient of Powers property.

\frac{3^2}{3^7} = 3^{2-7} = 3^{-5}

3. Are the answers from #1 and #2 equal? Write them as a single statement.

\frac{1}{3^5} = 3^{-5}

4. Fill in the blanks. \frac{1}{a^m} = a- and \frac{1}{a^{-m}} = a-

\frac{1}{a^m} = a^{-m} and \frac{1}{a^{-m}} = a^m

From the two investigations above, we have learned two very important properties of exponents. First, anything to the zero power is one. Second, negative exponents indicate placement. If an exponent is negative, it needs to be moved from where it is to the numerator or denominator. We will investigate this property further in the Problem Set.

Example A

Simplify the following expressions. Your answer should only have positive exponents.

(a) \frac{5^2}{5^5}

(b) \frac{x^7 yz^{12}}{x^{12} yz^7}

(c) \frac{a^4 b^0}{a^8 b}

Solution: Use the two properties from above. An easy way to think about where the “leftover” exponents should go, is to look at the fraction and determine which exponent is greater. For example, in b , there are more x ’s in the denominator, so the leftover should go there.

(a) \frac{5^2}{5^5} = 5^{-3} = \frac{1}{5^3} = \frac{1}{125}

(b) \frac{x^7 yz^{12}}{x^{12} yz^7} = \frac{y^{1-1} z^{12-7}}{x^{12-7}} = \frac{y^0 z^5}{x^5} = \frac{z^5}{x^5}

(c) \frac{a^4 b^0}{a^8 b} = a^{4-8} b^{0-1} = a^{-4} b^{-1} = \frac{1}{a^4 b}

Alternate Method : Part c

\frac{a^4 b^0}{a^8 b} = \frac{1}{a^{8-4} b} = \frac{1}{a^4 b}

Example B

Simplify the expressions. Your answer should only have positive exponents.

(a) \frac{xy^5}{8y^{-3}}

(b) \frac{27 g^{-7} h^0}{18 g}

Solution: In these expressions, you will need to move the negative exponent to the numerator or denominator and then change it to a positive exponent to evaluate. Also, simplify any numerical fractions.

(a) \frac{xy^5}{8y^{-3}} = \frac{xy^5 y^3}{8} = \frac{xy^{5+3}}{8} = \frac{xy^8}{8}

(b) \frac{27 g^{-7} h^0}{18 g} = \frac{3}{2g^1 g^7} = \frac{3}{2g^{1+7}} = \frac{3}{2g^8}

Example C

Multiply the two fractions together and simplify. Your answer should only have positive exponents.

\frac{4x^{-2} y^5}{20x^8} \cdot \frac{-5x^6 y}{15y^{-9}}

Solution: The easiest way to approach this problem is to multiply the two fractions together first and then simplify.

\frac{4x^{-2} y^5}{20x^8} \cdot \frac{-5x^6 y}{15y^{-9}} = -\frac{20x^{-2+6} y^{5+1}}{300x^8 y^{-9}} = -\frac{x^{-2+6-8}y^{5+1+9}}{15} = -\frac{x^{-4} y^{15}}{15} = -\frac{y^{15}}{15x^4}

Intro Problem Revisit

Set each earthquake's magnitude up as an exponential expression and divide.

\frac{10^{8.5}}{10^{9.5}}\\= 10^{-1}\\= \frac{1}{10^1}\\= \frac{1}{10}

Therefore, the the size of the 1575 earthquake was \frac{1}{10} the 1960 one.

Guided Practice

Simplify the expressions.

1. \frac{8^6}{8^9}

2. \frac{3x^{10} y^2}{21x^7 y^{-4}}

3. \frac{2a^8 b^{-4}}{16a^{-5}} \cdot \frac{4^3 a^{-3} b^0}{a^4 b^7}

Answers

1. \frac{8^6}{8^9} = 8^{6-9} = \frac{1}{8^3} = \frac{1}{512}

2. \frac{3x^{10} y^2}{21x^7 y^{-4}} = \frac{x^{10-7} y^{2-(-4)}}{7} = \frac{x^3 y^6}{7}

3. \frac{2a^8 b^{-4}}{16a^{-5}} \cdot \frac{4^3 a^{-3} b^0}{a^4 b^7} = \frac{128a^{8-3} b^{-4}}{16a^{-5+4} b^7} = \frac{8a^{5+1}}{b^{7+4}} = \frac{8a^6}{b^{11}}

Vocabulary

Zero Exponent Property
a^0 = 1, a \neq 0
Negative Exponent Property
\frac{1}{a^m} = a^{-m} and \frac{1}{a^{-m}} = a^m, a \neq 0

Practice

Simplify the following expressions. Answers cannot have negative exponents.

  1. \frac{8^2}{8^4}
  2. \frac{x^6}{x^{15}}
  3. \frac{7^{-3}}{7^{-2}}
  4. \frac{y^{-9}}{y^{10}}
  5. \frac{x^0 y^5}{xy^7}
  6. \frac{a^{-1} b^8}{a^5 b^7}
  7. \frac{14c^{10} d^{-4}}{21c^6 d^{-3}}
  8. \frac{8g^0 h}{30g^{-9} h^2}
  9. \frac{5x^4}{10y^{-2}} \cdot \frac{y^7 x}{x^{-1} y}
  10. \frac{g^9 h^5}{6gh^{12}} \cdot \frac{18h^3}{g^8}
  11. \frac{4a^{10} b^7}{12a^{-6}} \cdot \frac{9a^{-5} b^4}{20a^{11} b^{-8}}
  12. \frac{-g^8 h}{6g^{-8}} \cdot \frac{9g^{15} h^9}{-h^{11}}
  13. Rewrite the following exponential pattern with positive exponents: 5^{-4}, 5^{-3}, 5^{-2}, 5^{-1}, 5^0, 5^1, 5^2, 5^3, 5^4 .
  14. Evaluate each term in the pattern from #13.
  15. Fill in the blanks.

As the numbers increase, you ______________ the previous term by 5.

As the numbers decrease, you _____________ the previous term by 5.

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