Understanding Exponents - Algebra 2

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Question

Expand:

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Answer

To expand the exponent, we multiply the base by whatever the exponent is.

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Question

Evaluate: $\frac{3}{x^0+y^0}$

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Answer

Although we have two variables, we do know that a number raised to a zero power is one. Therefore:

$\frac{3}{x^0+y^0}=\frac{3}{1+1}=\frac{3}{2}$

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Question

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

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Answer

When dealing with exponents, always turn it into this form:

$\frac{1}{x^n}$ represents the base of the exponent, and is the power in a positive value.

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

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Question

Evaluate $(-8)^{-2}$

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Answer

When dealing with exponents, always turn it into this form:

$\frac{1}{x^n}$ represents the base of the exponent, and is the power in a positive value.

$\frac{1}{(-8)^{2}}=\frac{1}{64}$ It is important to keep the paranthesis as we are squaring which makes our answer.

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Question

Evaluate $(-9)^{-3}$

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Answer

When dealing with exponents, always turn it into this form:

$\frac{1}{x^n}$ represents the base of the exponent, and is the power in a positive value.

$\frac{1}{(-9)^{3}}=-\frac{1}{729}$ Our answer is negative because we have an odd exponent.

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Question

What is $5^{-2}$ the same as?

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Answer

While a positive exponent says how many times to multiply by a number, a negative exponent says how many times to divide by the number.

To solve for negative exponents, just calculate the reciprocal.

$5^{-2}=\frac{1}{5^{2}}=\frac{1}{25}$

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Question

Solve: $32^{-\frac{1}{5}}$

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Answer

To evaluate a negative exponent, convert the exponent to positive by taking the inverse.

$32^{-\frac{1}{5}}= \frac{1}{32^{\frac{1}{5}}}=\frac{1}{2}$

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Question

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

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Answer

When dealing with exponents, always turn it into this form:

$\frac{1}{x^n}$ represents the base of the exponent, and is the power in a positive value.

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

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Question

Convert the exponent to radical notation.

$x^{\frac{3}{7}}$

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Answer

Remember that exponents in the denominator refer to the root of the term, while exponents in the numerator can be treated normally.

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

$x^{\frac{3}{7}}=\sqrt[7]{x^3}$

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Question

Simplify:

$[x^{1/2}]^{7/3}$

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Answer

$[x^{m}]^{n}=x^{m\times n}$

$x^{1/2\times 7/3}=x^{7/6}$

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Question

Evaluate $-\frac{1}{6}^{-3}$ .

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Answer

When dealing with exponents, always turn it into this form:

$\frac{1}{x^n}$ represents the base of the exponent, and is the power in a positive value.

$\frac{1}{(-\frac{1}{6})^{3}}=\frac{1}{-\frac{1}{216}}=-216$

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Question

Simplify the expression:

$\small (16^{\frac{1}{2}})(256^{\frac{3}{4}})$

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Answer

Remember that fraction exponents are the same as radicals.

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

$256^{\frac{3}{4}}=\sqrt[4]{256^3}=64$

A shortcut would be to express the terms as exponents and look for opportunities to cancel.

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

$\small 256^{\frac{3}{4}}=(4^4)^{\frac{3}{4}}=4^3=64$

Either method, we then need to multiply to two terms.

$\small (4)(64)=256$

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Question

Write the product of $\small a^\frac{3}{4}a^\frac{3}{8}a^\frac{5}{2}$ in radical form

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Answer

This problem relies on the key knowledge that $\small x^\frac{a}{b}=\sqrt[b]{x^a}$ and that the multiplying terms with exponents requires adding the exponents. Therefore, we can rewrite the expression thusly:

$\small a^\frac{3}{4}a^\frac{3}{8}a^\frac{5}{2}=a^{\frac{3}{4}+\frac{3}{8}+\frac{5}{2}}=a^{\frac{29}{8}}=\sqrt[8]{a^{29}}$

Therefore, $\small \sqrt[8]{a^{29}}$ is our final answer.

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Question

Evaluate the following expression:

$\bigg(\frac{27}{125}\bigg)^{\frac{1}{3}}$

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Answer

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

or

$\bigg(\frac{27}{125}\bigg)^{\frac{1}{3}}= \sqrt[3]{\frac{27}{125}}=\frac{\sqrt[3]{27}}{\sqrt[3]{125}}=\frac{3}{5}$

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Question

Simplify:

$\frac{x^{\frac{1}{2}}y^2}{x^{\frac{-3}{2}}y^4}$

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Answer

Keep in mind that when you are dividing exponents with the same base, you will want to subtract the exponent found in the denominator from the exponent found in the numerator.

To find the exponent for , subtract the denominator's exponent from the numerator's exponent.

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

To find the exponent for , subtract the denominator's exponent from the numerator's exponent.

Since the exponent is negative, you will want to put the in the denominator in order to make it positive.

So then,

$\frac{x^{\frac{1}{2}}y^2}{x^{\frac{-3}{2}}y^4}=\frac{x^2}{y^2}$

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Question

Find the value of $9^{\frac{3}{2}}$ .

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Answer

When you have a number or value with a fractional exponent,

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

or

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

So then,

$9^{\frac{3}{2}}=\sqrt[2]{9^3}=\sqrt{729}=27$

$9^{\frac{3}{2}}=(\sqrt9)^3=3^3=27$

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Question

Find the value of $64^{\frac{4}{3}}$

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Answer

When you have a number or value with a fractional exponent,

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

or

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

So then,

$64^{\frac{4}{3}}=\sqrt[3]{64^4}=\sqrt[3]{2^{24}}=2^8=256$

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Question

Simplify:

$(27t^3)^\frac{5}{3}$

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Answer

When exponents are raised to another exponent, you will need to multiply the exponents together.

When you have a number or value with a fractional exponent,

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

or

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

So,

$27^{\frac{5}{3}}=\sqrt[3]{27^5}=\sqrt[3]{3^{15}}=3^5=243$

$(t^3)^{\frac{5}{3}}=t^{3\times\frac{5}{3}}=t^{\frac{15}{3}}=t^5$

$(27t^3)^\frac{5}{3}=243t^5$

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Question

Simplify:

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

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Answer

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

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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Question

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

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Answer

When dealing with exponents, always turn it into this form:

$\frac{1}{x^n}$ represents the base of the exponent, and is the power in a positive value.

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

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