Exponents - Math

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Question

Solve for :

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Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

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Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for :

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Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

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Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for :

Tap to reveal answer

Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

Tap to reveal answer

Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for :

Tap to reveal answer

Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

Tap to reveal answer

Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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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

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

Which of the following is equivalent to $64^{\frac{1}{2}}$ ?

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Answer

By definition, a number raised to the $\frac{1}{2}$ power is the same as the square root of that number.

Since the square root of 64 is 8, 8 is our solution.

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Question

Solve for :

$(x+5)^{-3} = -1$

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Answer

Raise both sides of the equation to the inverse power of to cancel the exponent on the left hand side of the equation.

$\rightarrow ((x+5)^{-3})^{-\frac{1}{3}} = (-1)^{-\frac{1}{3}}$

$\rightarrow x+5 = -1$

Subtract from both sides:

$\rightarrow (x+5) - 5 = (-1)-5$

$\rightarrow x = -6$

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Question

Which of the following is equivalent to $3^{-2}$ ?

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Answer

By definition,

$b^{-x} = \frac{1}{b^{x}}$ .

In our problem, and .

Then, we have $\frac{1}{3^{2}} = \frac{1}{9}$ .

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Question

Simplify the expression:

$(3x^4y^2)(4xy^2)^{-3}$

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Answer

Begin by distributing the exponent through the parentheses. The power rule dictates that an exponent raised to another exponent means that the two exponents are multiplied:

$(3x^{4}y^2)(4xy^2)^{-3}= (3xy^2)(4^{-3}x^{-3}y^{-6})$

Any negative exponents can be converted to positive exponents in the denominator of a fraction:

$\frac{3x^4y^2}{64x^3y^6}$

The like terms can be simplified by subtracting the power of the denominator from the power of the numerator:

$\frac{3x^4y^2}{64x^3y^6}=\frac{3}{64}x^{4-3}y^{2-6}=\frac{3}{64}xy^{-4}$

$\frac{3x}{64y^4}$

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Question

Order the following from least to greatest:

$25^{100}$

$2^{300}$

$3^{500}$

$4^{400}$

$2^{600}$

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Answer

In order to solve this problem, each of the answer choices needs to be simplified.

Instead of simplifying completely, make all terms into a form such that they have 100 as the exponent. Then they can be easily compared.

, , , and .

Thus, ordering from least to greatest: .

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Question

What is the largest positive integer, , such that is a factor of ?

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Answer

. Thus, is equal to 16.

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Question

$\log_{x}27e^{3}=3$

Solve for .

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Answer

First, set up the equation: . Simplifying this result gives .

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Question

$y=\frac{x-3}{x^2-12}$

What are the x-intercepts of this equation?

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Answer

To find the x-intercepts, set the numerator equal to zero.

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Question

Simplify the following expression.

$\frac{4x^3y^8z^2}{8x^6y^2z^4}$

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Answer

When dividing with exponents, the exponent in the denominator is subtracted from the exponent in the numerator. For example: $\frac{x^3}{x^6}= x^{3-6}=x^{-3}=\frac{1}{x^3}$ .

In our problem, each term can be treated in this manner. Remember that a negative exponent can be moved to the denominator.

$\frac{4x^3y^8z^2}{8x^6y^2z^4}=\frac{4}{8}x^{-3}y^6z^{-2}=\frac{4y^6}{8x^3z^2}$

Now, simplifly the numerals.

$\frac{4y^6}{8x^3z^2}=\frac{y^6}{2x^3z^2}$

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Question

Solve for :

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Answer

Rewrite each side of the equation to only use a base 2:

$2^{2y} = 2^{5}*2^{2y-x}$

$2^{2y} = 2^{5+2y-x}$

The only way this equation can be true is if the exponents are equal.

So:

The on each side cancel, and moving the to the left side, we get:

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