Exponents - High School Math

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Question

Simplify the expression:

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

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:

$25^{100}$

$2^{300}$

$3^{500}$

$4^{400}$

$2^{600}$

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 ?

Answer

. Thus, is equal to 16.

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Question

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

Solve for .

Answer

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

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Question

Find the -intercept(s) of $y=\frac{x^3+2x+8}{x^2-36}$ .

Answer

To find the -intercept, set in the equation and solve.

$y=\frac{x^3+2x+8}{x^2-36}$

$y=\frac{(0)^3+2(0)+8}{(0)^2-36}$

$y=\frac{(0)+(0)+8}{(0)-36}$

$y=\frac{8}{-36}$

$y=-\frac{2}{9}$

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Question

Find the -intercept(s) of $y=\frac{x^3+2x+8}{x^2-36}$ .

Answer

To find the -intercept(s) of $y=\frac{x^3+2x+8}{x^2-36}$ , we need to set the numerator equal to zero.

That means .

The best way to solve for a funky equation like this is to graph it in your calculator and calculate the roots. The result is $x\approx -1.67$ .

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Question

Find the -intercept(s) of $y=\frac{x^2+9x+8}{x^2-44}$ .

Answer

To find the -intercept(s) of $y=\frac{x^2+9x+8}{x^2-44}$ , we need to set the numerator equal to zero and solve.

First, notice that can be factored into . Now set that equal to zero: .

Since we have two sets in parentheses, there are two separate values that can cause our equation to equal zero: one where and one where .

Solve for each value:

and

.

Therefore there are two -interecpts: and .

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Question

Find the -intercept(s) of $y=\frac{x^2+9x+8}{x^2-44}$ .

Answer

To find the -intercept(s) of $y=\frac{x^2+9x+8}{x^2-44}$ , set the value equal to zero and solve.

$y=\frac{x^2+9x+8}{x^2-44}$

$y=\frac{(0)^2+9(0)+8}{(0)^2-44}$

$y=\frac{(0)+(0)+8}{(0)-44}$

$y=\frac{8}{-44}$

$y=-\frac{2}{11}$

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Question

Simplify the following expression.

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

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 :

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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Question

Simplify the expression:

$x^3 + \frac{2x^2}{x^{-1}}$

Answer

First simplify the second term, and then combine the two:

$x^3 + \frac{2x^2}{x^{-1}}$

$= x^3 + 2x^{2-(-1)} = x^3 + 2x^3 = 3x^3$

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Question

Simplify the following expression.

$2^{3} \cdot 2^{6}$

Answer

We are given: $2^{3} \cdot 2^{6}$ .

Recall that when we are multiplying exponents with the same base, we keep the base the same and add the exponents.

Thus, we have $2^{3 + 6} = 2^{9}$ .

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Question

Simplify the following expression.

$\frac{2^{9}}{2^{11}}$

Answer

Recall that when we are dividing exponents with the same base, we keep the base the same and subtract the exponents.

Thus, we have $\frac{2^{9}}{2^{11}} = 2^{9 - 11} = 2^{-2}$ .

We also recall that for negative exponents,

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

Thus, $2^{-2} = \frac{1}{2^{2}}$ .

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Question

Simplify the following exponent expression:

$(\frac{a^{\frac{-3}{2}}}{3^6b^{\frac{-2}{3}}})^{\frac{-1}{2}}$

Answer

Begin by rearranging the terms in the numerator and denominator so that the exponents are positive:

$(\frac{a^{\frac{-3}{2}}}{3^6b^{\frac{-2}{3}}})^{\frac{-1}{2}}$

$(\frac{3^6b^{\frac{-2}{3}}}{a^{\frac{-3}{2}}})^{\frac{1}{2}}$

$(\frac{3^6a^{\frac{3}{2}}}{b^{\frac{2}{3}}})^{\frac{1}{2}}$

Multiply the exponents:

$(\frac{3^3a^{\frac{3}{4}}}{b^{\frac{1}{3}}})$

Simplify:

$\frac{27a^{\frac{3}{4}}}{b^{\frac{1}{3}}}$

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Question

Solve the equation for .

$\small 9^x=3^6$

Answer

Begin by recognizing that both sides of the equation have a root term of .

$\small 9^x=3^6$

Using the power rule, we can set the exponents equal to each other.

$3^{(2*x)}=3^6$

$\small 2x=6$

$\small x=3$

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Question

Solve the equation for .

$\small 3^{2x}=81$

Answer

Begin by recognizing that both sides of the equation have the same root term, .

$\small 3^{2x}=81$

$\small 3^{2x}=9^2$

$3^{2x}=(3^2)^2$

We can use the power rule to combine exponents.

$3^{2x}=3^4$

Set the exponents equal to each other.

$\small x=2$

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Question

Which value for satisfies the equation $4^x = \frac{192}{x}$ ?

Answer

is the only choice from those given that satisfies the equation. Substition of for gives:

$4^3 = \frac{192}{3} \rightarrow 64 = 64$

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Question

Solve for :

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 :

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 (nearest hundredth):

$8^{y} = 100$

Answer

Take the common logarithm of both sides and solve for :

$8^{y} = 100$

$\log 8^{y} = \log 100$

$y \log 8 =2$

$y =\frac{2}{\log 8} \approx \frac{2}{0.9031} \approx 2.21$

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