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Algebra II : Multiplying and Dividing Exponents

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Example Question #181 : Multiplying And Dividing Exponents

Multiply: $(3^2)^{-3} \cdot 27$

Possible Answers:

$\frac{1}{27}$

$\frac{1}{54}$

$\frac{1}{3}$

$\frac{1}{18}$

Correct answer:

$\frac{1}{27}$

Explanation:

The powers of the term enclosed by the parentheses can be multiplied by the power rule.

$(3^2)^{-3} \cdot 27 = 3^{-6}\cdot 27$

We can rewrite as three cubed.

$3^{-6}\cdot 27 = 3^{-6}\cdot 3^3$

Add the powers using the additive rule.

$3^{-6+3} = 3^{-3} = \frac{1}{3^3} = \frac{1}{27}$

The answer is: $\frac{1}{27}$

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Example Question #182 : Multiplying And Dividing Exponents

Divide the exponents: $\frac{3^{-13}}{3^{-3}}\div 9^{-4}$

Possible Answers:

$\frac{4}{9}$

$\frac{1}{2}$

$\frac{1}{9}$

$\frac{1}{27}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{9}$

Explanation:

Evaluate the first fraction. Since both bases on the numerator and denominator are alike, we can use the subtraction rule of exponents in order to simplify the exponents.

$\frac{3^{-13}}{3^{-3}} = 3^{-13-(-3)} = 3^{-10}$

Rewrite the expression.

$\frac{ 3^{-10}}{9^{-4}}$

Change the base of the denominator to 3^2 .

$\frac{ 3^{-10}}{9^{-4}} = \frac{ 3^{-10}}{3^{2(-4)}} = \frac{ 3^{-10}}{3^{-8}}$

Simplify by using the subtraction rule.

$\frac{ 3^{-10}}{3^{-8}} = 3^{-10-(-8)} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

The answer is: $\frac{1}{9}$

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Example Question #183 : Multiplying And Dividing Exponents

Divide the exponents. $\frac{12}{12^{-1}}\div 3^{-1}$

Possible Answers:

432

$\frac{1}{3}$

243

Correct answer:

432

Explanation:

The first fraction can be rewritten as:

$\frac{12}{12^{-1}} = 12^{1-(-1)} = 12^{1+1} = 12^2 = 144$

The second term is the same as:

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

Divide the two numbers.

$144\div \frac{1}{3} = 144 \times \frac{3}{1} = 432$

The answer is: 432

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Example Question #184 : Multiplying And Dividing Exponents

Evaluate: $3(9)(\frac{1}{27})^{50}$

Possible Answers:

$\frac{1}{27^{49}}$

$\frac{1}{27^{51}}$

$\frac{1}{3^{53}}$

$\frac{1}{3^{101}}$

$\frac{1}{3^{51}}$

Correct answer:

$\frac{1}{27^{49}}$

Explanation:

Multiply the integers together.

$3(9)(\frac{1}{27})^{50}=27(\frac{1}{27})^{50}$

We can rewrite the fraction to base 27 using a negative exponent.

$27(27)^{-1(50)} = 27^1 \cdot 27^{-50}$

Recall the additive property of exponents. We can sum the exponents when similar bases are multiplied.

$27^{1+(-50)} = 27^{-49} = \frac{1}{27^{49}}$

The answer is: $\frac{1}{27^{49}}$

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Example Question #185 : Multiplying And Dividing Exponents

Divide: $\frac{x^{-18}}{x^{-30}} \div x^{-6}$

Possible Answers:

$x^{18}$

$\frac{1}{x^{55}}$

$\frac{1}{x^6}$

$\frac{1}{x^{42}}$

x^6

Correct answer:

$x^{18}$

Explanation:

Solve the first term. The fraction of this expression can be rewritten using the subtraction rule of exponents.

$\frac{x^{-18}}{x^{-30}} = x^{-18-(-30)} = x^{-18+30} = x^{12}$

Divide this by $x^{-6}$ .

$\frac{x^{12}}{x^{-6}}$

Repeat the process.

$x^{12-(-6)} = x^{18}$

The answer is: $x^{18}$

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Example Question #3641 : Algebra Ii

Divide: $\frac{8^{16}}{4^{24}}$

Possible Answers:

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{32}$

$\frac{1}{256}$

Correct answer:

Explanation:

Convert the bases so that the numerator and denominators have the common base. We can convert both bases to base 2.

Rewrite the bases in terms of the new base using exponents.

$\frac{8^{16}}{4^{24}} = \frac{2^{3(16)}}{2^{2(24)}} = \frac{2^{48}}{2^{48}} =1$

The answer is:

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Example Question #187 : Multiplying And Dividing Exponents

Simplify the following expression using multiplication and division of exponents to its simplest form.

2*x^8*x^9

Possible Answers:

$x^{17}$

$2x^{17}$

$2x^{72}$

$2x^{18}$

$x^{72}$

Correct answer:

$2x^{17}$

Explanation:

In order to multiply exponents with the same base of the form:

$a^{x}*a^{y}$ you must add the two exponents together, where a is the base and and are the exponents. When the base in this case "" is an expression such as you must check to see if they have the same base.

If they do $a^x*a^y=a^{(x+y)}$

In the case of the problem $2 x^{8}*x^{9}$

both x^8 and x^9 share the same base of x so you're able to add the exponents getting thereby getting $2x^{8+9}$ which equals = $2x^{17}$ , which is your final answer.

This can be checked by dividing $2x^{17}$ by either 2x^8 or x^9

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Algebra II : Multiplying and Dividing Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #181 : Multiplying And Dividing Exponents

Example Question #182 : Multiplying And Dividing Exponents

Example Question #183 : Multiplying And Dividing Exponents

Example Question #184 : Multiplying And Dividing Exponents

Example Question #185 : Multiplying And Dividing Exponents

Example Question #3641 : Algebra Ii

Example Question #187 : Multiplying And Dividing Exponents

All Algebra II Resources

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