High School Math : Multiplying and Dividing Exponents

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Simplifying Exponents

Simplify the following expression.

\displaystyle \frac{4x^3y^8z^2}{8x^6y^2z^4}

Possible Answers:

\displaystyle \frac{4y^4}{8x^2z^2}

\displaystyle \frac{y^4}{2x^2z^2}

\displaystyle \frac{4y^6}{8x^3z^2}

\displaystyle \frac{y^6}{2x^3z^2}

Correct answer:

\displaystyle \frac{y^6}{2x^3z^2}

Explanation:

When dividing with exponents, the exponent in the denominator is subtracted from the exponent in the numerator. For example: \displaystyle \frac{x^3}{x^6}= x^{3-6}=x^{-3}=\frac{1}{x^3}.

In our problem, each term can be treated in this manner. Remember that a negative exponent can be moved to the denominator.

\displaystyle \frac{4x^3y^8z^2}{8x^6y^2z^4}=\frac{4}{8}x^{-3}y^6z^{-2}=\frac{4y^6}{8x^3z^2}

Now, simplifly the numerals.

\displaystyle \frac{4y^6}{8x^3z^2}=\frac{y^6}{2x^3z^2}

Example Question #1 : Multiplying And Dividing Exponents

Simplify the following expression. 

 

\displaystyle 2^{3} \cdot 2^{6}

Possible Answers:

\displaystyle 4^{18}

\displaystyle 4^{9}

\displaystyle 2^{9}

\displaystyle 6^{6}

\displaystyle 2^{18}

Correct answer:

\displaystyle 2^{9}

Explanation:

We are given: \displaystyle 2^{3} \cdot 2^{6}

Recall that when we are multiplying exponents with the same base, we keep the base the same and add the exponents. 

Thus, we have \displaystyle 2^{3 + 6} = 2^{9}.

Example Question #2 : Simplifying Exponents

Simplify the following expression. 

 

\displaystyle \frac{2^{9}}{2^{11}}

Possible Answers:

\displaystyle \frac{18}{22}

\displaystyle 2^{2}

\displaystyle 2^{\frac{9}{11}}

\displaystyle \frac{1}{2^{2}}

\displaystyle \frac{9}{11}

Correct answer:

\displaystyle \frac{1}{2^{2}}

Explanation:

Recall that when we are dividing exponents with the same base, we keep the base the same and subtract the exponents. 

Thus, we have \displaystyle \frac{2^{9}}{2^{11}} = 2^{9 - 11} = 2^{-2}.

We also recall that for negative exponents,

\displaystyle b^{-x} = \frac{1}{b^{x}}.

Thus, \displaystyle 2^{-2} = \frac{1}{2^{2}}.

Example Question #9 : Exponents

Simplify the following exponent expression:

\displaystyle (\frac{a^{\frac{-3}{2}}}{3^6b^{\frac{-2}{3}}})^{\frac{-1}{2}}

Possible Answers:

\displaystyle \frac{a^{\frac{3}{4}}}{b^{\frac{1}{3}}}

\displaystyle \frac{3a^{\frac{3}{4}}}{b^{\frac{1}{3}}}

\displaystyle \frac{9a^{\frac{3}{4}}}{b^{\frac{1}{3}}}

\displaystyle \frac{27a^{\frac{3}{4}}}{b^{\frac{1}{3}}}

\displaystyle \frac{18a^{\frac{3}{4}}}{b^{\frac{1}{3}}}

Correct answer:

\displaystyle \frac{27a^{\frac{3}{4}}}{b^{\frac{1}{3}}}

Explanation:

Begin by rearranging the terms in the numerator and denominator so that the exponents are positive:

\displaystyle (\frac{a^{\frac{-3}{2}}}{3^6b^{\frac{-2}{3}}})^{\frac{-1}{2}}

\displaystyle (\frac{3^6b^{\frac{-2}{3}}}{a^{\frac{-3}{2}}})^{\frac{1}{2}}

\displaystyle (\frac{3^6a^{\frac{3}{2}}}{b^{\frac{2}{3}}})^{\frac{1}{2}}

Multiply the exponents:

\displaystyle (\frac{3^3a^{\frac{3}{4}}}{b^{\frac{1}{3}}})

Simplify:

\displaystyle \frac{27a^{\frac{3}{4}}}{b^{\frac{1}{3}}}

Example Question #2 : Multiplying And Dividing Exponents

Simplify the expression:

\displaystyle x^3 + \frac{2x^2}{x^{-1}}

Possible Answers:

\displaystyle x^3+2x

\displaystyle x^3+2

\displaystyle x^3+2x^2

\displaystyle 3x^3

\displaystyle x^3+\frac{2}{x^3}

Correct answer:

\displaystyle 3x^3

Explanation:

First simplify the second term, and then combine the two:

\displaystyle x^3 + \frac{2x^2}{x^{-1}}

\displaystyle = x^3 + 2x^{2-(-1)} = x^3 + 2x^3 = 3x^3

Example Question #1 : Simplifying Exponents

Solve for \displaystyle x\displaystyle 4^y = 32*2^2^y^-^x

Possible Answers:

\displaystyle x=3

\displaystyle x=5

\displaystyle x=0

\displaystyle x=-3

Cannot be determined from the given information.

Correct answer:

\displaystyle x=5

Explanation:

Rewrite each side of the equation to only use a base 2:

\displaystyle 2^{2y} = 2^{5}*2^{2y-x}

\displaystyle 2^{2y} = 2^{5+2y-x}

The only way this equation can be true is if the exponents are equal.

So:

\displaystyle 2y = 5+2y-x

The \displaystyle 2y on each side cancel, and moving the \displaystyle xto the left side, we get:

\displaystyle x=5

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