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Algebra II : Distributing Exponents (Power Rule)

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Example Question #71 : Distributing Exponents (Power Rule)

Simplify the exponent: $(2a^{120})^{5}$

Possible Answers:

$2a^{125}$

$32a^{125}$

$2a^{600}$

$16a^{600}$

$32a^{600}$

Correct answer:

$32a^{600}$

Explanation:

According to the property of the power rule for exponents,

$(x^a)^b = x^{a\times b}$

The exponents may be multiplied if the exponent is outside of the parentheses.

$(2a^{120})^{5} = 2^5 \cdot a^{120\times 5}$

The answer is: $32a^{600}$

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Example Question #72 : Distributing Exponents (Power Rule)

Simplify: $(a^2b^{25})^{20}$

Possible Answers:

$a^{22}b^{45}$

$a^{40}b^{400}$

$a^{22}b^{500}$

$\frac{a^{10}b^{5}}{2}$

$a^{40}b^{500}$

Correct answer:

$a^{40}b^{500}$

Explanation:

In order to simplify this, we will need to distribute the power of 20 across both powers inside the inner quantity.

$(a^2b^{25})^{20} = (a^{2\cdot 20}b^{25\cdot 20})$

Multiply the powers.

The answer is: $a^{40}b^{500}$

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Example Question #73 : Distributing Exponents (Power Rule)

Simplify: $(x^{35}\cdot x^{20})^{15}$

Possible Answers:

$x^{10500}$

$x^{70}$

$x^{715}$

$x^{825}$

$x^{80}$

Correct answer:

$x^{825}$

Explanation:

Simplify the inner quantity first.

When powers of the same base are multiplied together, the powers may be added.

$x^{35}\cdot x^{20} = x^{55}$

The expression becomes: $(x^{55})^{15}$

When the quantity of a power is raised to a certain power, the powers will need to be multiplied.

$(x^{55})^{15} =x^{55\times 15} = x^{825}$

The answer is: $x^{825}$

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Example Question #71 : Distributing Exponents (Power Rule)

Solve: $(2a^{300})^3$

Possible Answers:

$2a^{303}$

$8a^{900}$

$2a^{900}$

$6a^{303}$

$8a^{303}$

Correct answer:

$8a^{900}$

Explanation:

The term can be simplified by distributing the power with the term inside the parentheses.

$(2a^{300})^3 = 2^3\cdot a^{300\times 3} = 8a^{900}$

The answer is: $8a^{900}$

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Example Question #75 : Distributing Exponents (Power Rule)

Simplify the exponents: $[(a^{10})^5]^{20}$

Possible Answers:

$a^{35}$

$a^{300}$

$a^{3000}$

$a^{1000}$

$a^{70}$

Correct answer:

$a^{1000}$

Explanation:

Simplify the terms in the bracket first. When a quantity of a power is raised to the power, the exponents may be multiplied.

$[(a^{10})^5]^{20} =[(a^{10\times 5})]^{20} = [a^{50}]^{20}$

Repeat the process for the term in the bracket.

$a^{50\times 20}$

The answer is: $a^{1000}$

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Example Question #76 : Distributing Exponents (Power Rule)

Solve the following: $( 3^3\cdot 3^2)^{10}$

Possible Answers:

$6^{50}$

$3^{60}$

$9^{50}$

$3^{50}$

$3^{15}$

Correct answer:

$3^{50}$

Explanation:

Simplify the terms inside the parentheses.

$( 3^3\cdot 3^2) = (3^{3+2}) = 3^5$

$( 3^3\cdot 3^2)^{10} = (3^5)^{10}$

The quantity will be multiplied by itself ten times. This means that the powers here can be multiplied.

$(3^5)^{10} = 3^{5\times 10} = 3^{50}$

The answer is: $3^{50}$

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Example Question #77 : Distributing Exponents (Power Rule)

Simplify: $(6^{12})(\frac{1}{36})^{6}$

Possible Answers:

6^6

$6^{-144}$

$6^{-24}$

Correct answer:

Explanation:

Rewrite the second term so that it has the common base.

$\frac{1}{36} = 6^{-2}$

$(6^{12})(\frac{1}{36})^{6} = (6^{12})(6^{-2})^{6}$

Simplify the power outside of the parentheses by multiplication.

$(6^{12})(6^{-12})$

Since these exponents with the same bases are multiplied, the exponents can be added.

$(6^{12})(6^{-12})= 6^{12+(-12)} = 6^0=1$

The answer is:

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Example Question #78 : Distributing Exponents (Power Rule)

Simplify: $(9\cdot3^{12})^5$

Possible Answers:

$9^{60}$

$3^{120}$

$9^{50}$

$3^{70}$

$3^{80}$

Correct answer:

$3^{70}$

Explanation:

In order to simplify this, we will first need to simplify the inner term of the parentheses.

Rewrite the nine with base three.

9=3^2

The expression becomes:

$(3^2\cdot3^{12})^5$

Since the common bases of a certain power are multiplied, the exponents can be added.

$(3^2\cdot3^{12})^5=(3^{2+12})^5 = (3^{14})^5$

Now that we have an exponent outside of a quantity, we can multiply the exponents together.

$(3^{14})^5 = 3^{70}$

The answer is: $3^{70}$

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Example Question #79 : Distributing Exponents (Power Rule)

Solve: $(6^9)^{4}$

Possible Answers:

$1296^{4}$

$6^{36}$

$1296^{9}$

$24^{36}$

$6^{13}$

Correct answer:

$6^{36}$

Explanation:

This expression can be simplified by either writing out the quantities four time, or simply by just multiplying the powers together. Either method will give similar answers.

$(6^9)^{4}=(6^9)(6^9)(6^9)(6^9)= 6^{9\times 4} = 6^{36}$

Do not change the base at any point in the calculation.

The answer is: $6^{36}$

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Example Question #80 : Distributing Exponents (Power Rule)

Simplify: $(3^{25})^{25}$

Possible Answers:

$3^{500}$

$3^{675}$

$3^{50}$

$3^{625}$

$3^{250}$

Correct answer:

$3^{625}$

Explanation:

In order to solve this, use the distribution rule for exponents.

$(a^{X})^Y = a^{XY}$

Multiply the powers together.

$(3^{25})^{25} = 3^{25\times 25} = 3^{625}$

The answer is: $3^{625}$

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Algebra II : Distributing Exponents (Power Rule)

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #71 : Distributing Exponents (Power Rule)

Example Question #72 : Distributing Exponents (Power Rule)

Example Question #73 : Distributing Exponents (Power Rule)

Example Question #71 : Distributing Exponents (Power Rule)

Example Question #75 : Distributing Exponents (Power Rule)

Example Question #76 : Distributing Exponents (Power Rule)

Example Question #77 : Distributing Exponents (Power Rule)

Example Question #78 : Distributing Exponents (Power Rule)

Example Question #79 : Distributing Exponents (Power Rule)

Example Question #80 : Distributing Exponents (Power Rule)

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