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Algebra II : Fractional Exponents

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Example Questions

Example Question #3311 : Algebra Ii

Evaluate $\frac{1}{4}^{-\frac{3}{4}}$

Possible Answers:

$2\sqrt[3]{4}$

$\frac{\sqrt[4]{4}}{4}$

$2\sqrt[4]{4}$

$-\frac{\sqrt[4]{4}}{4}$

$-2\sqrt[4]{4}$

Correct answer:

$2\sqrt[4]{4}$

Explanation:

When dealing with fractional exponents, we write as

$x^\frac{a}{b}=\sqrt[b]{x^a}$

which is the index of the radical, is the exponent raising base .

We evaluate negative exponents as

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

which is the positive exponent raising base .

Therefore

$\frac{1}{4}^{-\frac{3}{4}}=\frac{1}{\frac{1}{\sqrt[4]{4^3}}}=\sqrt[4]{64}$ .

We can factor out 2^4 or .

$\sqrt[4]{64}=\sqrt[4]{16}*\sqrt[4]{4}=2*\sqrt[4]{4}=2\sqrt[4]{4}$

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Example Question #61 : Fractional Exponents

Simplify, if possible: $(\frac{3}{4})^{\frac{4}{3}}$

Possible Answers:

$\frac{3\sqrt[3]{3}}{4}$

$\frac{3\sqrt[3]{6}}{8}$

$\frac{3\sqrt[3]{6}}{4}$

$\frac{2\sqrt[3]{6}}{3}$

$\textup{The expression is already simplified.}$

Correct answer:

$\frac{3\sqrt[3]{6}}{8}$

Explanation:

In order to simplify this, we will need to multiply both the numerator and denominator with the outside power according to the power rule of exponents.

$(\frac{3}{4})^{\frac{4}{3}} = \frac{3^\frac{4}{3}}{4^\frac{4}{3}}$

The numerator of the power represents the power raised to. The denominator of the power represents the root of the radical. Rewrite the fractions.

$\frac{3^\frac{4}{3}}{4^\frac{4}{3}} = \frac{(\sqrt[3]{3})^4}{(\sqrt[3]{4})^4} = \frac{\sqrt[3]{3}\cdot \sqrt[3]{3}\cdot \sqrt[3]{3}\cdot \sqrt[3]{3}}{\sqrt[3]{4}\cdot \sqrt[3]{4}\cdot \sqrt[3]{4}\cdot \sqrt[3]{4}}$

Recall that multiplying the radicals in the third root three times will leave the integer by itself.

The expression becomes:

$\frac{\sqrt[3]{3}\cdot \sqrt[3]{3}\cdot \sqrt[3]{3}\cdot \sqrt[3]{3}}{\sqrt[3]{4}\cdot \sqrt[3]{4}\cdot \sqrt[3]{4}\cdot \sqrt[3]{4}}= \frac{3 \sqrt[3]{3}}{4 \sqrt[3]{4}}$

Rationalize the denominator by multiplying the top and bottom by $\sqrt[3]{4}$ twice in order to eliminate the cube root radical denominator.

$\frac{3 \sqrt[3]{3}}{4 \sqrt[3]{4}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}}= \frac{3 \cdot \sqrt[3]{48}}{4(4)} = \frac{3 \sqrt[3]{48}}{16}$

Simplify $\sqrt[3]{48}$ .

$\sqrt[3]{48} = \sqrt[3]{8\cdot 6} = \sqrt[3]{8} \cdot \sqrt[3]{6} = 2 \sqrt[3]{6}$

Replace the new term and reduce the fraction. The fraction becomes:

$\frac{3 \sqrt[3]{48}}{16} = \frac{3 \cdot 2 \sqrt[3]{6}}{16} = \frac{3\sqrt[3]{6}}{8}$

The answer is: $\frac{3\sqrt[3]{6}}{8}$

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Example Question #3313 : Algebra Ii

Simplify: $(\frac{2}{3})^{\frac{5}{2}}$

Possible Answers:

$\frac{11\sqrt6}{34}$

$\frac{2\sqrt6}{13}$

$\frac{7\sqrt5}{18}$

$\frac{4\sqrt6}{27}$

$\textup{This expression cannot be simplified.}$

Correct answer:

$\frac{4\sqrt6}{27}$

Explanation:

Use the product rule of exponents to simplify this term.

$(\frac{2}{3})^{\frac{5}{2}} = (\frac{2^{\frac{5}{2}}}{3^{\frac{5}{2}}})$

Rewrite this using radicals. The numerator represents the power that the radical is raised to. The denominator represents the root.

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

Multiply the terms together. A radical multiplied by itself will be the integer inside the radical. The terms become:

$\frac{2\cdot 2 \cdot\sqrt2}{3\cdot3\cdot\sqrt3} = \frac{4\sqrt2}{9\sqrt{3}}$

Rationalize the denominator. Multiply the top and bottom by square root three.

$\frac{4\sqrt2}{9\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt6}{9\times 3}$

The answer is: $\frac{4\sqrt6}{27}$

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Example Question #3314 : Algebra Ii

Which of the following is similar to $(3)^{\frac{7}{9}}$ ?

Possible Answers:

$(\frac{1}{3})^7$

$\frac{3^5}{9^7}$

$( \sqrt[7]{3})^9$

$(\frac{1}{3})^{\frac{7}{9}}$

$( \sqrt[9]{3})^7$

Correct answer:

$( \sqrt[9]{3})^7$

Explanation:

The fractional exponent will include both the power and the root. The numerator will represent the power that the quantity is raised to, and the denominator represents the root of the term.

Rewrite the expression in radical form.

$(3)^{\frac{7}{9}} =( \sqrt[9]{3})^7$

The answer is: $( \sqrt[9]{3})^7$

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Example Question #3315 : Algebra Ii

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

Possible Answers:

$\frac{\sqrt2}{3}$

$\frac{\sqrt6}{6}$

$\frac{\sqrt3}{9}$

$\frac{\sqrt3}{3}$

$\frac{\sqrt3}{2}$

Correct answer:

$\frac{\sqrt3}{3}$

Explanation:

Rewrite the half power with a radical.

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

Split the radical as two radicals.

$\sqrt{\frac{1}{3}} = \frac{\sqrt1}{\sqrt3}$

Rationalize the denominator. Multiply the top and bottom by square root three.

$\frac{\sqrt1}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3} =\frac{\sqrt3}{3}$

The answer is: $\frac{\sqrt3}{3}$

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Example Question #181 : Exponents

Evaluate: $[-16^{\frac{3}{2}}]^{\frac{1}{2}}$

Possible Answers:

$\textup{The answer does not exist.}$

-8i

Correct answer:

Explanation:

In order to solve this, we will need to rewrite the inner term as a radical.

$[-16^{\frac{3}{2}}]^{\frac{1}{2}} = [-(\sqrt{16})^3] ^{\frac{1}{2}}$

Simplify the inner term.

$[-64]^{\frac{1}{2}} = \sqrt{-64}$

The answer is:

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Example Question #182 : Exponents

Solve: $289^{\frac{3}{2}}-27^{\frac{2}{3}}$

Possible Answers:

4904

1825

1000

262

196

Correct answer:

4904

Explanation:

We can rewrite both terms using the radicals. The denominator of a fractional exponent is the index of the root. The numerator of the fraction is the power of the quantity.

Rewrite the terms.

$289^{\frac{3}{2}}-27^{\frac{2}{3}} =( \sqrt{289})^3 - (\sqrt[3]{27})^2$

Simplify the radicals and solve.

$289^{\frac{3}{2}}-27^{\frac{2}{3}} =(17)^3 - (3)^2 = 4913 -9 = 4904$

The answer is: 4904

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Example Question #3318 : Algebra Ii

Simplify:

$\frac{(x^2)(x^{\frac{1}{2}}y^9)}{x^{-\frac{4}{3}}y^5}$

Possible Answers:

$\frac{y^4}{x^{\frac{23}{6}}}$

$\frac{x^{\frac{23}{6}}}{y^4}$

$x^{\frac{8}{3}}y^9$

$x^\frac{23}{6}y^4$

Correct answer:

$x^\frac{23}{6}y^4$

Explanation:

Start by simplifying the numerator. Since two terms with the same base are being multiplied, add the exponents.

$\frac{(x^2)(x^{\frac{1}{2}}y^9)}{x^{-\frac{4}{3}}y^5}=\frac{x^{\frac{5}{2}}y^9}{x^{-\frac{4}{3}}y^5}$

Now, when terms with the same bases are divided, subtract the exponent from the denominator from the exponent in the numerator.

The exponent for is

$\frac{5}{2}-(-\frac{4}{3})=\frac{23}{6}$

The exponent for is

9-5=4

So then,

$\frac{(x^2)(x^{\frac{1}{2}}y^9)}{x^{-\frac{4}{3}}y^5}=\frac{x^{\frac{5}{2}}y^9}{x^{-\frac{4}{3}}y^5}=x^{\frac{23}{6}}y^4$

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Algebra II : Fractional Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3311 : Algebra Ii

Example Question #61 : Fractional Exponents

Example Question #3313 : Algebra Ii

Example Question #3314 : Algebra Ii

Example Question #3315 : Algebra Ii

Example Question #181 : Exponents

Example Question #182 : Exponents

Example Question #3318 : Algebra Ii

All Algebra II Resources

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