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Algebra II : Simplifying Exponents

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Example Questions

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Example Question #114 : Distributing Exponents (Power Rule)

Solve: $(8^{4})^{15}$

Possible Answers:

$8^{19}$

$2^{170}$

$4^{80}$

$2^{60}$

$8^{60}$

Correct answer:

$8^{60}$

Explanation:

Use the power rule of exponents to simplify this expression.

$(x^{Y})^Z = x^{YZ}$

Follow this rule to simplify the exponents.

$(8^{4})^{15} = 8^{4\times 15} = 8^{60}$

The answer is: $8^{60}$

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Example Question #115 : Distributing Exponents (Power Rule)

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

Possible Answers:

$\frac{1}{64}$

$\frac{1}{8}$

$\frac{1}{32}$

Correct answer:

$\frac{1}{8}$

Explanation:

We can use the power rule of exponents to simplify this expression.

$(x^{Y})^Z =x^{YZ}$

$[(2^{-3})^{-2}]^{-\frac{1}{2}} = 2^{(-3)\cdot (-2 )\cdot( -\frac{1}{2})}$

The expression becomes:

$2^{-3}$

We can rewrite this as a fraction.

$2^{-3} = \frac{1}{2^3}=\frac{1}{8}$

The answer is: $\frac{1}{8}$

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Example Question #116 : Distributing Exponents (Power Rule)

Simplify: $b(b^{-3})^{-10}$

Possible Answers:

$\frac{1}{b^{14}}$

$b^{14}$

$b^{31}$

$b^{29}$

$\frac{1}{b^{12}}$

Correct answer:

$b^{31}$

Explanation:

Use the product rule of exponents to simplify the terms in parentheses.

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

$b(b^{-3})^{-10} = b(b^{-3\times -10}) = b(b^{30})$

This expression is the same as: $b^1 \cdot b^{30}$

The answer is: $b^{31}$

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Example Question #117 : Distributing Exponents (Power Rule)

Solve: (5x^3)^4

Possible Answers:

$5x^{7}$

$625x^{12}$

$625x^{7}$

$5x^{12}$

$20x^{12}$

Correct answer:

$625x^{12}$

Explanation:

In order to solve this expression, we will need to use the power rule of exponents.

$(a^{x})^y = a^{xy}$

We can distribute the outer exponent to the exponents inside the parentheses. Do not ignore the five, which equals 5^1 .

$(5x^3)^4 = (5^{1\times 4}\cdot x^{3\times 4}) = 5^4 \cdot x^{12}$

The answer is: $625x^{12}$

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Example Question #301 : Simplifying Exponents

Solve: $\frac{1}{2}(64^{5})^{10}$

Possible Answers:

$4^{99}$

$2^{299}$

$4^{25}$

$2^{49}$

$4^{13}$

Correct answer:

$2^{299}$

Explanation:

In order to solve this, we can change this expression to base two.

The fraction can be rewritten as a negative exponent.

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

The inner term in the parentheses is equal to:

64= 2^6

Rewrite the expression.

$\frac{1}{2}(64^{5})^{10} = 2^{-1} (2^{6(5)})^{(10)}$

Using the product rule of exponents, we can multiply all the powers together.

$2^{-1} (2^{6\times 5\times 10})= 2^{-1} (2^{300}) = 2^{-1} \cdot (2^{300})$

Now that we are multiplying two common bases with exponents, we can add the powers.

$2^{-1+300} = 2^{299}$

The answer is: $2^{299}$

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Example Question #119 : Distributing Exponents (Power Rule)

Solve: 5[(5^3)^4]^5

Possible Answers:

$5^{18}$

$5^{61}$

$5^{13}$

$5^{12}$

$5^{60}$

Correct answer:

$5^{61}$

Explanation:

The exponents that are surrounded by the brackets and parentheses can be multiplied based on the product rule for exponents.

$5[(5^3)^4]^5 = 5^1\cdot [5^{3\times 4\times 5}] = 5^1\cdot 5^{60}$

Now that we have bases similar, we can use the additive rule of exponents to combine both terms.

The answer is: $5^{61}$

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Example Question #120 : Distributing Exponents (Power Rule)

Solve: $[(3^{-3})^2]^{\frac{1}{2}}$

Possible Answers:

$\frac{1}{27}$

$\frac{1}{9}$

Correct answer:

$\frac{1}{27}$

Explanation:

According to the product rule of exponents, we can multiply the two powers together in the parentheses.

$(x^B)^C = x^{B\times C}$

This means that:

$(3^{-3})^2 = 3^{-6}$

Replace this term in the bracket.

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

Repeat the process for these exponents.

$[3^{-6}]^{\frac{1}{2}} = 3^{-6\times \frac{1}{2}} = 3^{-3}$

This negative exponent can be simplified as a fraction.

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

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

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Example Question #301 : Simplifying Exponents

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

Possible Answers:

$\frac{1}{10}$

$\frac{1}{32}$

$\frac{1}{64}$

$\frac{1}{12}$

$\frac{1}{128}$

Correct answer:

$\frac{1}{64}$

Explanation:

In order to simplify this term, we will need to use the power rule of the exponents.

$(x^A)^B = x^{AB}$

Apply this rule for the given problem.

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

Simplify the terms.

$(\frac{1}{2})^{6} =(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) = \frac{1}{64}$

The answer is: $\frac{1}{64}$

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Algebra II : Simplifying Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #114 : Distributing Exponents (Power Rule)

Example Question #115 : Distributing Exponents (Power Rule)

Example Question #116 : Distributing Exponents (Power Rule)

Example Question #117 : Distributing Exponents (Power Rule)

Example Question #301 : Simplifying Exponents

Example Question #119 : Distributing Exponents (Power Rule)

Example Question #120 : Distributing Exponents (Power Rule)

Example Question #301 : Simplifying Exponents

All Algebra II Resources

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