Properties of Exponents in Algebraic Expressions Practice Questions — ACT

Question 1Question →

For all non-zero real numbers $a$ and $b$ , the expression $(a^2 b^k)^3$ is equivalent to $a^6 b^{15}$ . What is the value of the integer $k$ ?

Show answer & explanation

Answer: 5

Answer

The value of the integer

k

is 5.

To find the value of

k

, we simplify the expression

(a^2 b^k)^3

using exponent rules. According to the power of a product property,

(xy)^z = x^z y^z

, so

(a^2 b^k)^3 = (a^2)^3 (b^k)^3

. Next, applying the power of a power property,

(x^y)^z = x^{yz}

, we get

a^{2 \cdot 3} b^{k \cdot 3} = a^6 b^{3k}

. Since this expression is equivalent to

a^6 b^{15}

, we set the exponents of

b

equal to each other:

3k = 15

. Dividing both sides by 3 yields

k = 5

.

Step-by-Step Solution

1

Apply the power of a product rule to the expression

(a^2 b^k)^3

.

(a^2)^3 \cdot (b^k)^3

The power of a product rule states that

(xy)^z = x^z y^z

.

2

Apply the power of a power rule to simplify the exponents.

a^6 b^{3k}

The power of a power rule states that

(x^y)^z = x^{y \cdot z}

, so

(a^2)^3 = a^{2 \cdot 3} = a^6

and

(b^k)^3 = b^{3k}

.

3

Set the exponent of

b

in

a^6 b^{3k}

equal to the exponent of

b

in the equivalent expression

a^6 b^{15}

.

3k = 15

Since the expressions are equivalent for all non-zero real numbers, the exponents of like bases must be equal.

4

Solve the linear equation for

k

.

k = 5

Dividing both sides of

3k = 15

by 3 isolates the variable

k

.

Key Concept

Properties of exponents, specifically the power of a product rule

(xy)^z = x^z y^z

and the power of a power rule

(x^y)^z = x^{y \cdot z}

.

Estimated Time:45s

Question 2Question →

If the expression $\frac{(p^3 q^{-2})^{-1}}{(p^{-1} q^2)^3}$ is simplified to the form $p^x q^y$ for all non-zero real numbers $p$ and $q$ , what is the value of $2x - y$ ?

Show answer & explanation

Answer: 4

Answer

The correct answer is 4.

Applying the power of a power rule to the numerator yields

(p^3 q^{-2})^{-1} = p^{-3} q^2

. Applying the same rule to the denominator yields

(p^{-1} q^2)^3 = p^{-3} q^6

. Dividing the terms by subtracting exponents gives

p^{-3 - (-3)} q^{2-6} = p^0 q^{-4}

. Thus,

x = 0

and

y = -4

. Evaluating

2x - y

gives

2(0) - (-4) = 4

.

Step-by-Step Solution

1

Simplify the numerator using the power of a power property, which states that

(a^m)^n = a^{mn}

.

(p^3 q^{-2})^{-1} = p^{3 \cdot (-1)} q^{-2 \cdot (-1)} = p^{-3} q^2

This distributes the exponent of

-1

to both factors inside the parentheses in the numerator.

2

Simplify the denominator using the power of a power property, which states that

(a^m)^n = a^{mn}

.

(p^{-1} q^2)^3 = p^{-1 \cdot 3} q^{2 \cdot 3} = p^{-3} q^6

This distributes the exponent of

3

to both factors inside the parentheses in the denominator.

3

Combine the simplified numerator and denominator using the quotient property of exponents, which states that

\frac{a^m}{a^n} = a^{m-n}

.

\frac{p^{-3} q^2}{p^{-3} q^6} = p^{-3 - (-3)} q^{2 - 6} = p^0 q^{-4}

This simplifies division by subtracting the exponent in the denominator from the exponent in the numerator for each base.

4

Identify the values of

x

and

y

from the simplified form

p^0 q^{-4}

, and evaluate the final expression

2x - y

.

x = 0

and

y = -4

, so

2(0) - (-4) = 4

This substitutes the values of the exponents into the target algebraic expression to find the final numerical answer.

Key Concept

Properties of exponents including power of a power and quotient properties

Alternative Method

Alternatively, you can write the terms with positive exponents first:

\frac{(p^3 q^{-2})^{-1}}{(p^{-1} q^2)^3} = \frac{(p^{-1} q^2)^3}{(p^3 q^{-2})^1} = \frac{p^{-3} q^6}{p^3 q^{-2}} = p^{-3-3} q^{6-(-2)} = p^{-6} q^8

, but this expression is equivalent to the original expression only if inverted properly. Direct distribution is less prone to inversion errors.

Estimated Time:1m 30s

Question 3Question →

For all non-zero real numbers $x$ and $y$ , the algebraic expression $\left(\frac{x^{a/2} y^{-1}}{x^{-2/3} y^{b/3}}\right)^{-6}$ is equivalent to $\frac{y^{12}}{x^{16}}$ , where $a$ and $b$ are integers. What is the value of $a + b$ ?

-14

1

7

9

11

Show answer & explanation

Answer: 7

Answer

7

The correct answer is 7. Simplifying the expression inside the parentheses using the quotient rule yields

x^{\frac{a}{2} + \frac{2}{3}} y^{-1 - \frac{b}{3}}

. Raising this expression to the power of

-6

results in

x^{-3a - 4} y^{6 + 2b}

. Equating this to

x^{-16} y^{12}

gives

a = 4

and

b = 3

, so

a + b = 7

.

Step-by-Step Solution

1

Simplify the expression inside the parentheses using the quotient rule for exponents,

\frac{z^m}{z^n} = z^{m-n}

.

x^{\frac{a}{2} - \left(-\frac{2}{3}\right)} y^{-1 - \frac{b}{3}} = x^{\frac{a}{2} + \frac{2}{3}} y^{-1 - \frac{b}{3}}

To combine the base

x

and base

y

terms inside the parentheses before applying the outer power.

2

Apply the outer exponent of

-6

to the simplified expression using the power of a power rule,

(z^m)^n = z^{mn}

.

x^{-6\left(\frac{a}{2} + \frac{2}{3}\right)} y^{-6\left(-1 - \frac{b}{3}\right)} = x^{-3a - 4} y^{6 + 2b}

To distribute the negative power of

-6

to each factor in the product.

3

Equate the resulting exponents to the exponents of the given equivalent expression,

\frac{y^{12}}{x^{16}} = x^{-16} y^{12}

, and solve the equations for

a

and

b

.

For

x

:

-3a - 4 = -16 \implies -3a = -12 \implies a = 4

. For

y

:

6 + 2b = 12 \implies 2b = 6 \implies b = 3

.

Since the expressions are equivalent for all non-zero values of

x

and

y

, their respective exponents must be equal.

4

Calculate the sum of

a

and

b

.

a + b = 4 + 3 = 7

To find the final requested value.

Key Concept

Properties of exponents in algebraic expressions including the quotient rule, power rule, and operations with fractional/negative exponents.

Question 4Question →

For $x \neq 0$ , the expression $\frac{(x^2)^5}{x^k}$ simplifies to $x^6$ . What is the value of the exponent $k$ ?

Show answer & explanation

Answer: 4

Answer

The value of the exponent

k

is 4.

Applying the power of a power rule to the numerator yields

(x^2)^5 = x^{10}

. Next, applying the quotient rule to divide

x^{10}

by

x^k

yields

x^{10-k}

. Setting this equal to the simplified term

x^6

leads to the exponent equation

10 - k = 6

. Solving this equation gives the final result

k = 4

.

Step-by-Step Solution

1

Simplify the numerator expression

(x^2)^5

x^{10}

Multiply the exponents when raising a power to another power:

(x^a)^b = x^{ab}

.

2

Simplify the division of the two exponential expressions

\frac{x^{10}}{x^k}

x^{10-k}

Subtract the exponent of the denominator from the exponent of the numerator:

\frac{x^a}{x^b} = x^{a-b}

.

3

Solve the linear equation for

k

using the target exponent 6

k = 4

Equating the exponent

10-k

to

6

gives

10-k=6

. Subtracting 10 from both sides gives

-k = -4

, so

k = 4

.

Key Concept

Applying properties of exponents in algebraic expressions, specifically the power of a power rule and the quotient rule.

Question 5Question →

If the algebraic expression $\frac{(x^a y^2)^{-3} (x^2 y^b)^2}{(x^{-3} y^{-1})^{-2}}$ simplifies to $x^4 y^4$ for all non-zero real numbers $x$ and $y$ , where $a$ and $b$ are integers, what is the value of $a + b$ ?

Show answer & explanation

Answer: 4

Answer

The value of

a + b

is

4

.

Applying exponent rules to the expression

\frac{(x^a y^2)^{-3} (x^2 y^b)^2}{(x^{-3} y^{-1})^{-2}}

yields

\frac{x^{-3a}y^{-6} \cdot x^4 y^{2b}}{x^6 y^2} = \frac{x^{-3a+4} y^{2b-6}}{x^6 y^2} = x^{-3a-2} y^{2b-8}

. Equating these exponents to the target expression

x^4 y^4

gives the system

-3a - 2 = 4

and

2b - 8 = 4

. Solving these equations gives

a = -2

and

b = 6

, resulting in a sum of

a + b = 4

.

Step-by-Step Solution

1

Simplify the terms in the numerator.

(x^a y^2)^{-3} = x^{-3a} y^{-6}

and

(x^2 y^b)^2 = x^4 y^{2b}

Apply the power of a product rule:

(u^m v^n)^p = u^{mp} v^{np}

.

2

Multiply the simplified terms in the numerator.

x^{-3a+4} y^{2b-6}

Apply the product rule of exponents by adding exponents of like bases:

u^m \cdot u^n = u^{m+n}

.

3

Simplify the denominator.

(x^{-3} y^{-1})^{-2} = x^6 y^2

Apply the power of a product rule.

4

Divide the numerator by the denominator.

x^{-3a-2} y^{2b-8}

Apply the quotient rule of exponents by subtracting denominator exponents from numerator exponents:

\frac{u^m}{u^n} = u^{m-n}

.

5

Set up equations by equating the simplified exponents to the exponents in the target expression

x^4 y^4

.

-3a - 2 = 4

and

2b - 8 = 4

For the expressions to be equivalent for all non-zero real numbers, the corresponding exponents of

x

and

y

must be equal.

6

Solve the linear equations for the integer constants

a

and

b

.

a = -2

and

b = 6

Isolate the variables:

-3a = 6 \implies a = -2

, and

2b = 12 \implies b = 6

.

7

Find the sum of

a

and

b

.

4

Add the values of the constants:

-2 + 6 = 4

.

Key Concept

Properties of exponents (product, quotient, and power rules) in multi-step algebraic simplification

Question 6Question →

Which of the following expressions is equivalent to $\frac{12x^8}{4x^2}$ for all $x \neq 0$ ?

8x^4

8x^6

3x^4

3x^6

3x^{10}

Show answer & explanation

Answer: $3x^6$

Answer

The expression

3x^6

To simplify the expression, divide the coefficients first:

\frac{12}{4} = 3

. Then, apply the quotient rule of exponents to the variable terms, which states that for any non-zero base,

\frac{x^a}{x^b} = x^{a-b}

. Subtracting the exponents gives

x^{8-2} = x^6

. Combining these results yields the correct expression

3x^6

.

Step-by-Step Solution

1

Divide the numerical coefficients of the terms.

\frac{12}{4} = 3

When simplifying a fraction with algebraic terms, the coefficients are divided normally.

2

Apply the quotient rule of exponents to simplify the variable terms.

\frac{x^8}{x^2} = x^{8-2} = x^6

According to the quotient rule of exponents, when dividing expressions with the same base, subtract the exponent in the denominator from the exponent in the numerator.

3

Multiply the simplified coefficient and variable results together.

3x^6

Combining the divided coefficient and the simplified variable expression gives the final simplified result.

Key Concept

Quotient Rule of Exponents

Alternative Method

Alternatively, you can expand the exponent terms in the numerator and denominator:

\frac{12 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x}{4 \cdot x \cdot x}

. Simplifying the coefficients gives

3

, and canceling two pairs of

x

from both the numerator and denominator leaves six factors of

x

in the numerator, which simplifies to

3x^6

.

Estimated Time:45s

Question 7Question →

For all non-zero real numbers $x$ and $y$ , which of the following is equivalent to the expression $\frac{\left( 2x^{-1} y^2 + \frac{1}{2} x^{-1} y^2 \right)^{-3}}{(x^2 y^{-3})^{-2}}$ ?

\frac{5x^7}{2y^{12}}

\frac{8x^7}{125y^{12}}

\frac{65x^7}{8y^{12}}

\frac{8}{125x^{12}y^{36}}

\frac{x^7}{y^{12}}

Show answer & explanation

Answer: $\frac{8x^7}{125y^{12}}$

Answer

\frac{8x^7}{125y^{12}}

To find the equivalent expression, we first combine the like terms inside the parentheses in the numerator to get

\frac{5}{2} x^{-1} y^2

. Raising this product to the power of

-3

yields

\left(\frac{5}{2}\right)^{-3} (x^{-1})^{-3} (y^2)^{-3} = \frac{8}{125} x^3 y^{-6}

. Next, the denominator simplifies to

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

. Dividing the numerator by the denominator requires subtracting the exponents of like bases: for

x

, we have

3 - (-4) = 7

, and for

y

, we have

-6 - 6 = -12

. This results in

\frac{8}{125} x^7 y^{-12}

, which is equivalent to the correct expression

\frac{8x^7}{125y^{12}}

.

Step-by-Step Solution

1

Combine the like terms inside the parentheses in the numerator.

2x^{-1} y^2 + \frac{1}{2} x^{-1} y^2 = \left(2 + \frac{1}{2}\right) x^{-1} y^2 = \frac{5}{2} x^{-1} y^2

Before applying the outer negative exponent, it is mathematically simpler to combine the like terms inside the grouping.

2

Apply the power of

-3

to the term in the numerator.

\left(\frac{5}{2} x^{-1} y^2\right)^{-3} = \left(\frac{5}{2}\right)^{-3} (x^{-1})^{-3} (y^2)^{-3} = \frac{8}{125} x^3 y^{-6}

The power of a product rule

(ab)^n = a^n b^n

and the power of a power rule

(a^m)^n = a^{mn}

are applied to expand the term.

3

Simplify the denominator by applying the power of

-2

.

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

The power of a product rule is applied to the denominator to resolve the outer exponent.

4

Divide the simplified numerator by the simplified denominator using the quotient rule for exponents.

\frac{\frac{8}{125} x^3 y^{-6}}{x^{-4} y^6} = \frac{8}{125} x^{3 - (-4)} y^{-6 - 6} = \frac{8}{125} x^7 y^{-12} = \frac{8x^7}{125y^{12}}

The quotient rule

a^m / a^n = a^{m-n}

is used to subtract the exponents of the corresponding variables, and negative exponents are rewritten in the denominator.

Key Concept

Properties of Exponents in Algebraic Expressions

Alternative Method

Alternatively, you can write out all variables with positive exponents before simplifying. Rewrite the term in the numerator as

\frac{2y^2}{x} + \frac{y^2}{2x} = \frac{5y^2}{2x}

. Raising this to the

-3

power flips the fraction and cubes it, yielding

\left(\frac{2x}{5y^2}\right)^3 = \frac{8x^3}{125y^6}

. Simplifying the denominator yields

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

. Dividing the numerator by the denominator yields

\frac{8x^3}{125y^6} \div \frac{y^6}{x^4} = \frac{8x^3}{125y^6} \cdot \frac{x^4}{y^6} = \frac{8x^7}{125y^{12}}

.

Estimated Time:2m 0s

Question 8Question →

For all non-zero real numbers $x$ , the expression $x^4 \cdot (x^3)^k$ is equivalent to $x^{10}$ . What is the value of the integer $k$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

First, use the power of a power property to write

(x^3)^k

as

x^{3k}

. The expression then becomes

x^4 \cdot x^{3k}

. Next, use the product of powers property to combine the terms into

x^{4+3k}

. Since the expression is equivalent to

x^{10}

, set the exponents equal:

4 + 3k = 10

. Solving this equation gives

3k = 6

, which simplifies to

k = 2

.

Step-by-Step Solution

1

Apply the power of a power rule

(x^a)^b = x^{ab}

to simplify

(x^3)^k

.

x^{3k}

To raise a power to another power, multiply the exponents.

2

Apply the product of powers rule

x^a \cdot x^b = x^{a+b}

to combine the terms

x^4 \cdot x^{3k}

.

x^{4+3k}

When multiplying exponential terms with the same base, add their exponents.

3

Set the combined exponent

4+3k

equal to the target exponent

10

and solve for

k

.

k = 2

Since the bases are equal and non-zero, their exponents must be equal.

Key Concept

Properties of exponents in algebraic expressions (power of a power rule and product of powers rule)

Estimated Time:45s

Question 9Question →

For all non-zero real numbers $x$ and $y$ , the expression

\frac{(x^2 y^{-3})^{-2} (x^{-1} y^4)^3}{(x^3 y^{-2})^d}

can be written in the form $x^p y^q$ , where $p$ and $q$ are integers. If $q = 2p$ , what is the value of $d$ ?

Show answer & explanation

Answer: -4

Answer

-4

Applying the rules of exponents yields the simplified expression

x^{-7-3d} y^{18+2d}

. Setting the exponent of

y

equal to twice the exponent of

x

gives the equation

18+2d = 2(-7-3d)

, which solves to

d = -4

.

Step-by-Step Solution

1

Apply the power of a power rule to the terms in the numerator.

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

and

(x^{-1} y^4)^3 = x^{-3} y^{12}

To raise a power to another power, multiply the exponents:

(u^m)^n = u^{mn}

.

2

Multiply the simplified terms in the numerator together.

x^{-4} y^6 \cdot x^{-3} y^{12} = x^{-7} y^{18}

To multiply powers with the same base, add the exponents:

u^m \cdot u^n = u^{m+n}

.

3

Apply the power of a power rule to the denominator.

(x^3 y^{-2})^d = x^{3d} y^{-2d}

Distribute the exponent

d

to both variables inside the parentheses by multiplying the exponents.

4

Divide the numerator by the denominator.

\frac{x^{-7} y^{18}}{x^{3d} y^{-2d}} = x^{-7-3d} y^{18-(-2d)} = x^{-7-3d} y^{18+2d}

To divide powers with the same base, subtract the exponent of the denominator from the exponent of the numerator:

\frac{u^m}{u^n} = u^{m-n}

.

5

Set up the linear equation for

d

using

q = 2p

and solve.

18+2d = 2(-7-3d) \implies 18+2d = -14-6d \implies 8d = -32 \implies d = -4

The problem states the relationship between the final exponents is

q = 2p

, where

p = -7-3d

and

q = 18+2d

.

Key Concept

Properties of exponents (product, quotient, and power rules) combined with solving a linear equation.

Question 10Question →

For any positive real number $y$ , the expression $\frac{(y^3)^{1/2}}{y^{2/3}}$ is equivalent to which of the following?

y^{7/6}

y^{1/6}

y^{5/6}

y^{17/6}

y^{13/6}

Show answer & explanation

Answer: $y^{5/6}$

Answer

The simplified expression is

y^{5/6}

Applying the power of a power rule to the numerator gives

(y^3)^{1/2} = y^{3/2}

. Then, applying the quotient rule to divide by

y^{2/3}

requires subtracting the exponents:

\frac{3}{2} - \frac{2}{3} = \frac{9}{6} - \frac{4}{6} = \frac{5}{6}

. This results in the equivalent expression

y^{5/6}

.

Step-by-Step Solution

1

Apply the power of a power rule to the numerator

(y^3)^{1/2}

.

y^{3 \cdot 1/2} = y^{3/2}

When raising a power to a power, multiply the exponents.

2

Apply the quotient of powers rule to divide

y^{3/2}

by

y^{2/3}

.

y^{3/2 - 2/3}

When dividing powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.

3

Subtract the fractions in the exponent by finding a common denominator.

y^{9/6 - 4/6} = y^{5/6}

A common denominator for 2 and 3 is 6. Rewrite the fractions and subtract their numerators.

Key Concept

Properties of Exponents in Algebraic Expressions

Question 11Question →

If $x$ and $y$ are positive real numbers, the expression

\frac{(x^{1/2} + y^{1/2})^2 - (x^{1/2} - y^{1/2})^2}{(2x^{-1/4} y^{3/4})^{-2}}

can be simplified to the form $k y^p$ . What is the value of the sum $k + p$ ?

1

10

15

17

18

Show answer & explanation

Answer: 18

Answer

18

Expanding the numerator yields

4x^{1/2}y^{1/2}

. Simplifying the denominator using exponent rules yields

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

. Dividing the numerator by the denominator gives

\frac{4}{1/4} \cdot x^{1/2 - 1/2} \cdot y^{1/2 - (-3/2)} = 16y^2

, which corresponds to

k = 16

and

p = 2

. The sum is

16 + 2 = 18

.

Step-by-Step Solution

1

Expand and simplify the numerator.

4x^{1/2}y^{1/2}

Expand both squared binomials:

(x^{1/2} + y^{1/2})^2 = x + 2x^{1/2}y^{1/2} + y

and

(x^{1/2} - y^{1/2})^2 = x - 2x^{1/2}y^{1/2} + y

. Subtracting the second expression from the first yields

(x + 2x^{1/2}y^{1/2} + y) - (x - 2x^{1/2}y^{1/2} + y) = 4x^{1/2}y^{1/2}

.

2

Simplify the denominator using exponent rules.

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

Apply the power of a product rule

(ab)^n = a^n b^n

to distribute the exponent of

-2

:

(2x^{-1/4}y^{3/4})^{-2} = 2^{-2} \cdot (x^{-1/4})^{-2} \cdot (y^{3/4})^{-2}

. This simplifies to

\frac{1}{4} \cdot x^{(-1/4)(-2)} \cdot y^{(3/4)(-2)} = \frac{1}{4}x^{1/2}y^{-3/2}

.

3

Divide the simplified numerator by the simplified denominator.

16y^2

Divide the coefficients and subtract the exponents of like bases:

\frac{4x^{1/2}y^{1/2}}{\frac{1}{4}x^{1/2}y^{-3/2}} = \left(\frac{4}{1/4}\right) x^{1/2 - 1/2} y^{1/2 - (-3/2)} = 16 x^0 y^{1/2 + 3/2} = 16y^2

.

4

Identify the values of

k

and

p

and calculate their sum.

18

Comparing

16y^2

to the form

k y^p

gives

k = 16

and

p = 2

. Therefore, the sum is

k + p = 16 + 2 = 18

.

Key Concept

Properties of Exponents in Algebraic Expressions

Question 12Question →

If $a$ and $b$ are non-zero real numbers, which of the following expressions is equivalent to $\frac{a^4 b^{-2}}{(a^2 b^{-3})^2}$ ?

a^2 b^4

b^{-1}

b^{-8}

b^4

b^{-11}

Show answer & explanation

Answer: $b^4$

Answer

b^4

To simplify the expression, first apply the power of a product rule to the denominator:

(a^2 b^{-3})^2 = (a^2)^2 (b^{-3})^2 = a^4 b^{-6}

. Next, divide the numerator by the simplified denominator by subtracting the exponents of the corresponding bases:

\frac{a^4 b^{-2}}{a^4 b^{-6}} = a^{4-4} b^{-2 - (-6)} = a^0 b^4

. Since

a^0 = 1

for any non-zero real number

a

, the expression simplifies to

b^4

.

Step-by-Step Solution

1

Simplify the denominator using the power of a product and power of a power rules.

(a^2 b^{-3})^2 = (a^2)^2 \cdot (b^{-3})^2 = a^4 b^{-6}

When raising a product to a power, raise each factor to that power. When raising a power to a power, multiply the exponents.

2

Substitute the simplified denominator back into the original fraction.

\frac{a^4 b^{-2}}{a^4 b^{-6}}

To prepare the expression for division.

3

Divide the numerator by the denominator by subtracting the exponents of like bases.

a^{4-4} b^{-2 - (-6)} = a^0 b^4 = 1 \cdot b^4 = b^4

When dividing terms with the same base, subtract the exponent of the denominator from the exponent of the numerator.

Key Concept

Properties of exponents, specifically the power of a product rule, power of a power rule, and quotient rule.

Question 13Question →

For positive real values of $u$ and $v$ , the expression $\left( \frac{9u^{-1}v^2}{4u^3v^{-4}} \right)^{-1/2}$ can be simplified to which of the following?

\frac{2u^2v}{3}

\frac{2u^2}{3v^3}

\frac{3u^2}{2v^3}

\frac{2}{3uv^3}

\frac{4u^2}{9v^3}

Show answer & explanation

Answer: $\frac{2u^2}{3v^3}$

Answer

The expression is equivalent to

\frac{2u^2}{3v^3}

.

The correct answer is obtained by first simplifying the quotient inside the parenthesis to get

\frac{9}{4}u^{-4}v^6

. Then, raising each factor to the

-1/2

power yields

\left(\frac{9}{4}\right)^{-1/2} = \frac{2}{3}

,

(u^{-4})^{-1/2} = u^2

, and

(v^6)^{-1/2} = v^{-3}

. Combining these results and rewriting with positive exponents gives the simplified expression.

Step-by-Step Solution

1

Simplify the

u

terms inside the parenthesis using the quotient rule for exponents.

\frac{u^{-1}}{u^3} = u^{-1 - 3} = u^{-4}

The quotient rule states that

\frac{x^a}{x^b} = x^{a-b}

.

2

Simplify the

v

terms inside the parenthesis using the quotient rule for exponents.

\frac{v^2}{v^{-4}} = v^{2 - (-4)} = v^6

Subtracting a negative exponent is equivalent to adding its absolute value.

3

Apply the outer exponent of

-1/2

to the coefficient.

\left(\frac{9}{4}\right)^{-1/2} = \left(\frac{4}{9}\right)^{1/2} = \frac{2}{3}

A negative exponent represents taking the reciprocal of the base, and a fractional exponent of

1/2

represents the square root.

4

Apply the outer exponent of

-1/2

to the simplified variable terms using the power of a power rule.

(u^{-4})^{-1/2} = u^2

and

(v^6)^{-1/2} = v^{-3}

The power of a power rule states that

(x^a)^b = x^{ab}

.

5

Combine the simplified parts and rewrite the expression with positive exponents.

\frac{2}{3} u^2 v^{-3} = \frac{2u^2}{3v^3}

An expression with a negative exponent in the numerator can be moved to the denominator with a positive exponent.

Key Concept

Applying product, quotient, and power rules of exponents to algebraic expressions with negative and rational exponents.

Question 14Question →

The algebraic expression $\frac{(x^2)^a}{x^{-4}}$ is equivalent to $x^{10}$ for all non-zero real numbers $x$ . What is the value of $a$ ?

Show answer & explanation

Answer: 3

Answer

The correct answer is 3.

The correct value for

a

is

3

. Applying the power of a power rule to

(x^2)^a

yields

x^{2a}

. Then, applying the quotient rule to

\frac{x^{2a}}{x^{-4}}

yields

x^{2a - (-4)} = x^{2a+4}

. Equating the exponents gives

2a + 4 = 10

, which solves to

a = 3

.

Step-by-Step Solution

1

Apply the power of a power property to the numerator.

(x^2)^a = x^{2a}

When raising a power to another power, multiply the exponents:

(x^m)^n = x^{mn}

.

2

Apply the quotient property of exponents to simplify the fraction.

\frac{x^{2a}}{x^{-4}} = x^{2a - (-4)} = x^{2a + 4}

When dividing exponential expressions with the same base, subtract the exponent in the denominator from the exponent in the numerator:

\frac{x^m}{x^n} = x^{m-n}

.

3

Set the resulting exponent equal to the exponent of the equivalent expression and solve for

a

.

2a + 4 = 10 \implies 2a = 6 \implies a = 3

Since the bases are identical and the expressions are equivalent, their exponents must be equal.

Key Concept

Properties of Exponents in Algebraic Expressions

Question 15Question →

For all positive real numbers

x

and

y

, the expression

\frac{(3x^2 y^{-3})^3 \cdot (2x^{-1} y^2)^2}{(6x^3 y^{-2})^2}

can be simplified to the form

A x^a y^b

, where

A

,

a

, and

b

are integers. What is the value of the sum

A + a + b

?

Show answer & explanation

Answer: 0

Answer

The value of the sum

A + a + b

is 0.

By applying the rules of exponents systematically, the expression simplifies to

3 x^{-2} y^{-1}

. Comparing this to

A x^a y^b

yields

A = 3

,

a = -2

, and

b = -1

. The sum is

3 + (-2) + (-1) = 0

.

Step-by-Step Solution

1

Simplify the first term in the numerator

27x^6y^{-9}

Apply the power of a product rule and power of a power rule to

(3x^2 y^{-3})^3

.

2

Simplify the second term in the numerator

4x^{-2}y^4

Apply the power of a product rule and power of a power rule to

(2x^{-1} y^2)^2

.

3

Multiply the simplified terms in the numerator together

108x^4y^{-5}

Multiply coefficients and add the exponents of like bases.

4

Simplify the denominator

36x^6y^{-4}

Apply the power of a product rule and power of a power rule to

(6x^3 y^{-2})^2

.

5

Divide the numerator by the denominator

3x^{-2}y^{-1}

Divide the coefficients and subtract the denominator exponents from the numerator exponents for like bases.

6

Sum the constants

A

,

a

, and

b

0

Identify

A = 3

,

a = -2

,

b = -1

from the expression

3x^{-2}y^{-1}

, and calculate

3 + (-2) + (-1) = 0

.

Key Concept

Properties of Exponents in Algebraic Expressions

Question 16Question →

If $a$ and $b$ represent positive real numbers, which of the following is an equivalent form of the expression below?

\frac{(a^{-1} + b^{-1})^{-2}}{ab}

\frac{a^2 + b^2}{ab}

\frac{(a+b)^2}{4ab}

\frac{ab}{(a+b)^2}

\frac{(a+b)^2}{ab}

\frac{1}{a^2 b^2 (a+b)^2}

Show answer & explanation

Answer: $\frac{ab}{(a+b)^2}$

Answer

The correct answer is the fraction with

ab

in the numerator and the square of the sum

(a+b)

in the denominator, which is \frac{ab}{(a+b)^2}.

The correct answer is found by first rewriting

a^{-1} + b^{-1}

as

\frac{1}{a} + \frac{1}{b}

, which simplifies to

\frac{a+b}{ab}

using a common denominator. Raising this to the power of

-2

gives

\frac{a^2 b^2}{(a+b)^2}

. Finally, dividing this expression by

ab

reduces the exponents of

a

and

b

by 1, resulting in

\frac{ab}{(a+b)^2}

.

Step-by-Step Solution

1

Rewrite the negative exponents in the numerator as reciprocals.

The expression inside the parentheses becomes

\frac{1}{a} + \frac{1}{b}

.

By the definition of negative exponents,

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

.

2

Find a common denominator to add the fractions inside the parentheses.

\frac{1}{a} + \frac{1}{b} = \frac{b}{ab} + \frac{a}{ab} = \frac{a+b}{ab}

.

To add fractions, they must share a common denominator, which is the product of

a

and

b

.

3

Apply the negative power of

-2

to the simplified fraction.

\left(\frac{a+b}{ab}\right)^{-2} = \left(\frac{ab}{a+b}\right)^2 = \frac{a^2 b^2}{(a+b)^2}

.

An expression raised to a negative exponent is equal to the reciprocal of the expression raised to the positive exponent.

4

Divide the result by the denominator

ab

.

\frac{a^2 b^2}{(a+b)^2} \cdot \frac{1}{ab} = \frac{ab}{(a+b)^2}

.

Dividing by a term is equivalent to multiplying by its reciprocal, and applying the exponent quotient rule simplifies

\frac{a^2 b^2}{ab}

to

ab

.

Key Concept

Properties of exponents including negative power rules, quotient rules, and algebraic fraction addition.

Alternative Method

An alternative method is to substitute small positive integer values for

a

and

b

. Let

a = 1

and

b = 2

. The expression evaluates to

\frac{(1^{-1} + 2^{-1})^{-2}}{1 \cdot 2} = \frac{(1 + 0.5)^{-2}}{2} = \frac{1.5^{-2}}{2} = \frac{(3/2)^{-2}}{2} = \frac{4/9}{2} = \frac{2}{9}

. Evaluating the correct expression with these values yields

\frac{1 \cdot 2}{(1+2)^2} = \frac{2}{9}

, which matches.

Estimated Time:2m 0s

Question 17Question →

For all non-zero real numbers

y

, the expression

\frac{(y^3)^2 \cdot (y^{-4})^a}{y^5}

is equivalent to

y^{-7}

. What is the value of

a

?

2

-3

-4

\frac{1}{12}

\frac{3}{2}

Show answer & explanation

Answer: $2$

Answer

The correct value of

a

is 2.

Applying the exponent rules systematically allows us to simplify the expression. First,

(y^3)^2

becomes

y^6

and

(y^{-4})^a

becomes

y^{-4a}

using the power of a power rule. Second, we combine the terms in the numerator using the product rule to get

y^{6-4a}

. Third, we divide by

y^5

using the quotient rule to obtain

y^{6-4a-5} = y^{1-4a}

. Setting this equal to the target expression

y^{-7}

gives the equation

1-4a = -7

. Solving for

a

gives

-4a = -8

, which simplifies to

2

.

Step-by-Step Solution

1

Apply the power of a power rule,

(x^m)^n = x^{mn}

, to the exponential terms in the numerator.

(y^3)^2 = y^6

and

(y^{-4})^a = y^{-4a}

This simplifies nested exponent terms into single base terms.

2

Apply the product rule of exponents,

x^m \cdot x^n = x^{m+n}

, to combine the numerator terms.

y^6 \cdot y^{-4a} = y^{6-4a}

This simplifies the numerator to a single power of

y

.

3

Apply the quotient rule of exponents,

\frac{x^m}{x^n} = x^{m-n}

, to divide by the denominator.

\frac{y^{6-4a}}{y^5} = y^{(6-4a) - 5} = y^{1-4a}

This simplifies the entire rational expression into a single exponential expression.

4

Equate the simplified exponent to the exponent of the equivalent expression and solve the linear equation for

a

.

1-4a = -7 \implies -4a = -8 \implies a = 2

Since the bases are equal and non-zero, their exponents must be equal for the expressions to be equivalent.

Key Concept

Properties of Exponents in Algebraic Expressions

Estimated Time:1m 30s

Question 18Question →

If

x

and

y

are positive real numbers such that

(x^3 y^{-2})^k \cdot (x^{-1} y^4)^3 = x^6 y^6

what is the value of the exponent

k

?

Show answer & explanation

Answer: 3

Answer

The value of the exponent

k

is 3.

Applying the exponent rules simplifies the left side of the equation to

x^{3k-3}y^{12-2k}

. Equating the exponent of

x

to the right side gives

3k - 3 = 6

, which yields

k = 3

. This value is confirmed by equating the exponent of

y

, since

12 - 2(3) = 6

.

Step-by-Step Solution

1

Apply the power of a power rule

(a^m)^n = a^{mn}

to expand the terms in the expression.

(x^3 y^{-2})^k = x^{3k} y^{-2k}

and

(x^{-1} y^4)^3 = x^{-3} y^{12}

To remove the outer parentheses by multiplying the internal exponents of each variable by the outer exponent.

2

Multiply the terms together by applying the product rule for exponents,

a^m \cdot a^n = a^{m+n}

.

(x^{3k} y^{-2k})(x^{-3} y^{12}) = x^{3k-3} y^{12-2k}

To combine the like bases of

x

and

y

into a single simplified expression.

3

Set the exponents of like bases equal to the exponents on the right-hand side of the equation,

x^6 y^6

.

3k - 3 = 6

and

12 - 2k = 6

Since the bases are equal and non-zero, their respective exponents must also be equal.

4

Solve the linear equation

3k - 3 = 6

for

k

.

3k = 9 \implies k = 3

To determine the numerical value of the variable

k

.

5

Verify the solution by solving the second linear equation,

12 - 2k = 6

.

-2k = -6 \implies k = 3

To ensure consistency across both variable exponents in the expression.

Key Concept

Properties of Exponents in Algebraic Expressions

Question 19Question →

If the expression $\frac{(x^2 y^3)^4}{(x^{-1} y^2)^3}$ is written in the equivalent form $x^a y^b$ , what is the value of $a - b$ ?

Show answer & explanation

Answer: 5

Answer

5

Simplifying the numerator yields

x^8 y^{12}

and simplifying the denominator yields

x^{-3} y^6

. Dividing these expressions by subtracting the exponents of like bases results in

x^{8 - (-3)} y^{12 - 6} = x^{11} y^6

. Comparing this to the expression

x^a y^b

shows that

a = 11

and

b = 6

. The value of

a - b

is

11 - 6 = 5

.

Step-by-Step Solution

1

Simplify the numerator of the expression.

x^8 y^{12}

Apply the power of a power and power of a product properties of exponents:

(x^2 y^3)^4 = x^{2 \cdot 4} y^{3 \cdot 4}

.

2

Simplify the denominator of the expression.

x^{-3} y^6

Apply the power of a power and power of a product properties of exponents:

(x^{-1} y^2)^3 = x^{-1 \cdot 3} y^{2 \cdot 3}

.

3

Simplify the quotient by dividing the simplified numerator by the simplified denominator.

x^{11} y^6

Use the quotient property of exponents,

\frac{x^m}{x^n} = x^{m-n}

, to subtract the exponents:

8 - (-3) = 11

and

12 - 6 = 6

.

4

Identify the values of

a

and

b

and compute the difference

a - b

.

5

By comparing

x^{11} y^6

to the target form

x^a y^b

, we find

a = 11

and

b = 6

. Subtracting

b

from

a

yields

11 - 6 = 5

.

Key Concept

Properties of exponents including power of a power, power of a product, and quotient rules.

Question 20Question →

For all non-zero real numbers

a

,

b

, and

c

, the expression below is simplified:

\frac{(2a^2 b^{-1} c^3)^3}{4a^5 (b^2 c^{-2})^{-2}}

Which of the following is equivalent to this expression?

\frac{2ac^5}{b^7}

\frac{abc^5}{2}

2abc^5

4abc^5

2b^2c^{10}

Show answer & explanation

Answer: $2abc^5$

Answer

The expression is equivalent to

2abc^5

.

The correct answer is obtained by first simplifying the numerator and denominator using the power of a product and power of a power rules, then dividing the coefficients and subtracting the exponents of like bases. This yields the simplified expression

2abc^5

.

Step-by-Step Solution

1

Apply the power of a product rule to the numerator.

(2a^2 b^{-1} c^3)^3 = 2^3 \cdot (a^2)^3 \cdot (b^{-1})^3 \cdot (c^3)^3 = 8 a^6 b^{-3} c^9

Each factor inside the parentheses must be raised to the power of 3, multiplying the exponents of the variables.

2

Apply the power of a product rule to the denominator's parentheses.

4a^5 (b^2 c^{-2})^{-2} = 4a^5 \cdot (b^2)^{-2} \cdot (c^{-2})^{-2} = 4a^5 b^{-4} c^4

The terms inside the parentheses are raised to the power of -2, multiplying their exponents.

3

Divide the simplified numerator by the simplified denominator.

2a^1 b^1 c^5

, which is

2abc^5

Divide the coefficients (8 / 4 = 2) and subtract the exponents of the same bases:

a^{6-5} = a^1

,

b^{-3 - (-4)} = b^1

, and

c^{9-4} = c^5

.

Key Concept

Properties of Exponents in Algebraic Expressions

Alternative Method

Alternatively, you can rewrite the expression by eliminating negative exponents first. Convert

b^{-1}

to

\frac{1}{b}

,

b^2

to itself, and

c^{-2}

to

\frac{1}{c^2}

within the parentheses, apply the outer exponents, and then simplify the resulting complex fraction.

Estimated Time:1m 30s

Properties of Exponents in Algebraic Expressions

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