Exponents - Math

Card 0 of 212

Question

Simplify the expression:

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

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.

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

Compare your answer with the correct one above

Question

Convert the exponent to radical notation.

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

Answer

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

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

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

Compare your answer with the correct one above

Question

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

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.

Compare your answer with the correct one above

Question

Solve for :

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

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:

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

$\rightarrow x = -6$

Compare your answer with the correct one above

Question

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

Answer

By definition,

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

In our problem, and .

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

Compare your answer with the correct one above

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:

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

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

Compare your answer with the correct one above

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: .

Compare your answer with the correct one above

Question

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

Answer

. Thus, is equal to 16.

Compare your answer with the correct one above

Question

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

Solve for .

Answer

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

Compare your answer with the correct one above

Question

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

What are the x-intercepts of this equation?

Answer

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

Compare your answer with the correct one above

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.

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

Compare your answer with the correct one above

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:

Compare your answer with the correct one above

Question

Simplify the expression:

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

Answer

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

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

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

Compare your answer with the correct one above

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 $2^{3 + 6} = 2^{9}$ .

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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.

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

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

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

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

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

Compare your answer with the correct one above

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 $x\approx -1.67$ .

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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.

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

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

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

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

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

Compare your answer with the correct one above

Tap the card to reveal the answer