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Algebra II : Exponents

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Example Questions

Example Question #3233 : Algebra Ii

Evaluate: $7^{-1}-2(3)^{-1}$

Possible Answers:

$-\frac{1}{42}$

$-\frac{17}{21}$

$-\frac{3}{7}$

$-\frac{1}{21}$

$-\frac{11}{21}$

Correct answer:

$-\frac{11}{21}$

Explanation:

In order to evaluate this, we will first need to rewrite the negative exponents fractions.

$x^{-b} = \frac{1}{x^b}$

Rewrite each term.

$7^{-1}-2(3)^{-1} = \frac{1}{7} - 2(\frac{1}{3}) = \frac{1}{7}-\frac{2}{3}$

Determine the least common denominator by multiplying both denominators together.

$\frac{1}{7}-\frac{2}{3} = \frac{1(3)}{7(3)}-\frac{2(7)}{3(7)} = \frac{3}{21}-\frac{14}{21}$

Subtract the numerator.

The answer is: $-\frac{11}{21}$

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Example Question #3234 : Algebra Ii

Simplify: $2(3^{-2})-3(2^{-2})+6^{-1}$

Possible Answers:

$\frac{41}{36}$

$-\frac{13}{3}$

$-\frac{13}{36}$

$-\frac{27}{12}$

$\frac{7}{6}$

Correct answer:

$-\frac{13}{36}$

Explanation:

Rewrite all negative exponential into fractions.

$x^{-y} =\frac{1}{x^y}$

$2(3^{-2})-3(2^{-2})+6^{-1} = 2(\frac{1}{3^2})-3(\frac{1}{2^2})+\frac{1}{6}$

Simplify the terms.

$\frac{2}{9}-\frac{3}{4}+\frac{1}{6}$

Determine the least common factor by writing out the multiples of each denominator.

9-[9,18,27,36]

4-[4,8,12,16,20,24,28,32,36]

6-[6,12,18,24,30,36]

Convert each fraction to a common denominator of 36.

$\frac{2}{9}-\frac{3}{4}+\frac{1}{6}=\frac{2(4)}{9(4)}-\frac{3(9)}{4(9)}+\frac{1(6)}{6(6)}$

Simplify the fractions.

$\frac{8}{36}-\frac{27}{36}+\frac{6}{36} = -\frac{13}{36}$

The answer is: $-\frac{13}{36}$

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Example Question #101 : Understanding Exponents

Simplify: $\frac{(3^{-2})^{-1}}{2^{-1}}$

Possible Answers:

$\frac{2}{9}$

$\frac{1}{9}$

$\frac{1}{18}$

$\frac{1}{36}$

Correct answer:

Explanation:

Simplify the numerator by product of exponents.

$\frac{(3^{-2})^{-1}}{2^{-1}}=\frac{3^2}{2^{-1}}= \frac{9}{2^{-1}}$

Rewrite the negative exponent as a fraction.

$x^{-y}=\frac{1}{x^y}$

$\frac{9}{2^{-1}} = \frac{9}{\frac{1}{2}}= 9\div \frac{1}{2}$

Convert the division sign to a multiplication sign and take the reciprocal of the second term.

$9\times \frac{2}{1} =18$

The answer is:

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Example Question #3231 : Algebra Ii

Solve: $(\frac{1}{3})^{-3}$

Possible Answers:

$\frac{1}{27}$

$\frac{1}{9}$

$-\frac{1}{9}$

Correct answer:

Explanation:

In order to evaluate this expression, we will need to rewrite the negative exponent as a fraction.

$x^{-B}= \frac{1}{x^B}$

$(\frac{1}{3})^{-3} =\frac{1}{(\frac{1}{3})^3} = \frac{1}{(\frac{1}{3})(\frac{1}{3})(\frac{1}{3})}$

Simplify the denominator.

$\frac{1}{\frac{1}{27}} = 1\div \frac{1}{27} = 1\times \frac{27}{1} = 27$

The answer is:

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Example Question #3232 : Algebra Ii

Evaluate: $\frac{2^{-4}}{3^{-3}}$

Possible Answers:

$\frac{9}{8}$

$\frac{27}{16}$

$\frac{16}{27}$

$\frac{1}{432}$

$\frac{8}{27}$

Correct answer:

$\frac{27}{16}$

Explanation:

In order to simplify this expression, we will need to reconvert the negative exponents to fractions.

$x^{-y} = \frac{1}{x^y}$

Rewrite the fraction.

$\frac{2^{-4}}{3^{-3}} = \frac{\frac{1}{2^4}}{\frac{1}{3^3}} = \frac{\frac{1}{16}}{\frac{1}{27}}$

Rewrite the complex fraction using a division sign.

$\frac{\frac{1}{16}}{\frac{1}{27}} = \frac{1}{16}\div \frac{1}{27} = \frac{1}{16}\times \frac{27} {1}$

The answer is: $\frac{27}{16}$

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Example Question #3238 : Algebra Ii

Evaluate: $\frac{4^{-2}}{(\frac{1}{2})^{-3}}$

Possible Answers:

$\frac{1}{128}$

128

$\frac{1}{2}$

$\frac{1}{64}$

Correct answer:

$\frac{1}{128}$

Explanation:

We will need to convert both negative exponents into fractions.

$x^{-a} =\frac{1}{x^a}$

The expression becomes:

$\frac{4^{-2}}{(\frac{1}{2})^{-3}}=\frac{\frac{1}{4^{2}}}{\frac{1}{(\frac{1}{2})^{3}}}$

Evaluate each term.

$\frac{1}{4^{2}} =\frac{1}{16}$

$\frac{1}{(\frac{1}{2})^{3}} = \frac{1}{\frac{1}{8}} = 1\div \frac{1}{8} = 1\times \frac{8}{1}=8$

This means that:

$\frac{\frac{1}{4^{2}}}{\frac{1}{(\frac{1}{2})^{3}}} = \frac{1}{16} \div 8 = \frac{1}{16}\times \frac{1}{8}$

The answer is: $\frac{1}{128}$

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Example Question #3239 : Algebra Ii

Solve: $3(\frac{5}{3})^{-3}$

Possible Answers:

$\frac{125}{9}$

$\frac{125}{81}$

$\frac{9}{125}$

$\frac{81}{125}$

$\frac{1}{125}$

Correct answer:

$\frac{81}{125}$

Explanation:

Rewrite the negative exponential term as a fraction.

$x^{-y}=\frac{1}{x^y}$

Rewrite the expression.

$3(\frac{5}{3})^{-3} = 3\cdot \frac{1}{(\frac{5}{3})^3} = 3\cdot \frac{1}{(\frac{125}{27})}$

Simplify the second term.

$\frac{1}{(\frac{125}{27})} = 1\div \frac{125}{27} = 1\times \frac{27} {125}=\frac{27} {125}$

Replace the term.

$3\cdot \frac{1}{(\frac{125}{27})} = 3\cdot \frac{27}{125}$

Multiply the whole number with the numerator.

The answer is: $\frac{81}{125}$

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Example Question #101 : Exponents

Solve: $2(\frac{2}{7})^ {-2}$

Possible Answers:

$\frac{8}{7}$

$\frac{8}{49}$

$\frac{49}{2}$

$\frac{343}{4}$

$\frac{7}{2}$

Correct answer:

$\frac{49}{2}$

Explanation:

Solve by first rewriting the term with the negative exponent as a fraction.

$2(\frac{2}{7})^ {-2} = 2\cdot \frac{1}{(\frac{2}{7})^ {2}} = 2\cdot \frac{1}{\frac{4}{49}}$

Simplify the complex fraction.

$\frac{1}{\frac{4}{49}} = 1\div \frac{4}{49} =1 \times \frac{49}{4}=\frac{49}{4}$

The expression becomes:

$2\cdot \frac{1}{\frac{4}{49}} = 2\cdot \frac{49}{4} = \frac{49}{2}$

The answer is: $\frac{49}{2}$

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Example Question #3241 : Algebra Ii

Evaluate: $(\frac{3^{-2}}{6^{-1}})^{-2}$

Possible Answers:

$\frac{9}{4}$

2916

$\frac{1}{2916}$

$\frac{9}{16}$

$\frac{16}{9}$

Correct answer:

$\frac{9}{4}$

Explanation:

To evaluate a negative exponent, it is the reciprocal of the positive power.

$x^{-y} = \frac{1}{x^y}$

Evaluate each term.

$3^{-2} = \frac{1}{3^2} =\frac{1}{9}$

$6^{-1} = \frac{1}{6}$

Divide both fractions.

$(\frac{3^{-2}}{6^{-1}}) = \frac{1}{9}\div \frac{1}{6} = \frac{1}{9}\times \frac{6}{1} = \frac{2}{3}$

The original expression was: $(\frac{3^{-2}}{6^{-1}})^{-2}$

Replace the term and repeat the process for solving negative exponents.

$(\frac{3^{-2}}{6^{-1}})^{-2} = (\frac{2}{3})^{-2} = \frac{1}{(\frac{2}{3})^2} = \frac{1}{\frac{4}{9}}$

The answer is: $\frac{9}{4}$

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Example Question #101 : Exponents

Solve: $3^{-1}+3^{-2}-3^{-3}$

Possible Answers:

$\frac{13}{27}$

$\frac{11}{27}$

$\frac{7}{27}$

$-\frac{11}{27}$

$\frac{4}{27}$

Correct answer:

$\frac{11}{27}$

Explanation:

Convert all terms with negative exponents to fractional form.

$x^{-Y} = \frac{1}{x^Y}$

$3^{-1}+3^{-2}-3^{-3} =\frac{1}{3^{1}}+\frac{1}{3^2}-\frac{1}{3^3}$

$\frac{1}{3}+\frac{1}{9}-\frac{1}{27}$

Convert all fractions to a common denominator. Visually, the LCD can be seen as 27, since this value is divisible by all three terms.

$\frac{1(9)}{3(9)}+\frac{1(3)}{9(3)}-\frac{1}{27} = \frac{9}{27}+ \frac{3}{27}- \frac{1}{27}$

The answer is: $\frac{11}{27}$

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Algebra II : Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3233 : Algebra Ii

Example Question #3234 : Algebra Ii

Example Question #101 : Understanding Exponents

Example Question #3231 : Algebra Ii

Example Question #3232 : Algebra Ii

Example Question #3238 : Algebra Ii

Example Question #3239 : Algebra Ii

Example Question #101 : Exponents

Example Question #3241 : Algebra Ii

Example Question #101 : Exponents

All Algebra II Resources

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