Properties of Exponents in Algebraic Expressions Practice Questions — ACT — Page 2

If $x$ is a real number such that $\frac{(3^x)^4 \cdot 9^{2-x}}{27^{x-1}} = 81$ , what is the value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The value of

x

is 3.

Rewriting the bases in terms of 3, the expression becomes

\frac{3^{4x} \cdot 3^{4-2x}}{3^{3x-3}} = 3^4

. Combining the numerator using the product rule gives

3^{2x+4}

in the numerator. Dividing by the denominator using the quotient rule gives

3^{(2x+4)-(3x-3)} = 3^{-x+7}

. Equating this to

3^4

gives

-x + 7 = 4

, which yields

x = 3

.

Step-by-Step Solution

1

Express each base in terms of the common base 3

9 = 3^2

,

27 = 3^3

, and

81 = 3^4

Converting all terms to a common base allows the use of exponent rules to combine them.

2

Apply the power of a power rule

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

to rewrite each exponent

(3^x)^4 = 3^{4x}

,

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

, and

(3^3)^{x-1} = 3^{3x-3}

This simplifies individual terms by multiplying their exponents.

3

Apply the product rule of exponents

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

to combine the terms in the numerator

3^{4x} \cdot 3^{4-2x} = 3^{4x + 4 - 2x} = 3^{2x + 4}

Multiplying exponential terms with the same base is simplified by adding their exponents.

4

Apply the quotient rule of exponents

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

to simplify the fraction

3^{(2x+4) - (3x-3)} = 3^{-x+7}

Dividing exponential terms with the same base is simplified by subtracting the exponent in the denominator from the exponent in the numerator.

5

Equate the exponents of the simplified base 3 expression and base 3 representation of 81

-x + 7 = 4

, which solves to

x = 3

Since the bases are equal, the powers must be equal for the equation to hold true.

Key Concept

Properties of Exponents in Algebraic Expressions

Question 22Question →

For all non-zero real numbers

p

and

q

, which of the following expressions is equivalent to

\frac{(p^2 + q^{-1})^2 - p^4}{q^{-2}}?

1

2p^2 q^2 + 1

2p^2 q + 1

2p^2 q^{-1} + 1

2p^2 + 1

Show answer & explanation

Answer: $2p^2 q + 1$

Answer

2p^2 q + 1

Expanding the binomial in the numerator yields

p^4 + 2p^2 q^{-1} + q^{-2}

. After subtracting

p^4

, the numerator is left as

2p^2 q^{-1} + q^{-2}

. Dividing this numerator term-by-term by the denominator

q^{-2}

gives

\frac{2p^2 q^{-1}}{q^{-2}} + \frac{q^{-2}}{q^{-2}}

. Applying the quotient property of exponents to each term yields

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

.

Step-by-Step Solution

1

Expand the binomial in the numerator using the perfect square identity:

(a+b)^2 = a^2 + 2ab + b^2

.

(p^2 + q^{-1})^2 = (p^2)^2 + 2(p^2)(q^{-1}) + (q^{-1})^2 = p^4 + 2p^2 q^{-1} + q^{-2}

To begin simplifying the numerator, we must resolve the exponent outside the parentheses.

2

Subtract

p^4

from the expanded numerator expression.

(p^4 + 2p^2 q^{-1} + q^{-2}) - p^4 = 2p^2 q^{-1} + q^{-2}

This simplifies the numerator by combining the

p^4

terms.

3

Divide each term in the simplified numerator by the denominator

q^{-2}

.

\frac{2p^2 q^{-1}}{q^{-2}} + \frac{q^{-2}}{q^{-2}} = 2p^2 q^{-1 - (-2)} + q^{-2 - (-2)}

To divide a polynomial by a monomial, we distribute the division to each term of the polynomial.

4

Apply the quotient rule of exponents,

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

, and simplify the terms.

2p^2 q^1 + q^0 = 2p^2 q + 1

Subtracting the exponents of

q

in each term simplifies the division to its final form.

Key Concept

Properties of Exponents in Algebraic Expressions

Question 23Question →

If the expression

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

is equivalent to $x^2 y$ for all non-zero real numbers $x$ and $y$ , what is the value of the exponent $k$ ?

Show answer & explanation

Answer: 3

Answer

The value of the exponent

k

is 3.

Applying exponent rules simplifies the numerator of the expression to

x^5 y^4

. Dividing this by the denominator

x^k y^3

yields

x^{5-k} y

. Equating the exponent of

x

to 2 in the target expression

x^2 y

gives

5 - k = 2

, which solves to

k = 3

.

Step-by-Step Solution

1

Apply the power of a power rule

(a^m)^n = a^{m \cdot n}

to simplify each factor in the numerator.

(x^3 y^2)^3 = x^9 y^6

and

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

To raise a product to a power, raise each factor to that power by multiplying the exponents.

2

Multiply the two simplified factors in the numerator together by adding the exponents of like bases.

x^9 y^6 \cdot x^{-4} y^{-2} = x^5 y^4

When multiplying exponential expressions with the same base, add their exponents:

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

.

3

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

\frac{x^5 y^4}{x^k y^3} = x^{5-k} y

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

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

.

4

Set the exponent of

x

in the simplified expression equal to the exponent of

x

in the target expression

x^2 y

, and solve for

k

.

5 - k = 2 \implies k = 3

For the expressions to be equivalent for all non-zero real numbers, the corresponding exponents of like bases must be equal.

Key Concept

Properties of exponents (power of a power, product of powers, and quotient of powers rules)

Estimated Time:1m 30s

Question 24Question →

If $x$ and $y$ are positive real numbers, what is the simplified form of the expression below?

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

2x^{7/5}y

\frac{2y}{x^{1/6}}

2x^{11/6}y

2x^{7/6}y

8x^{11/6}y

Show answer & explanation

Answer: $2x^{11/6}y$

Answer

The simplified form of the expression is

2x^{11/6}y

.

The correct answer is

2x^{11/6}y

. First, simplify the constant coefficients to obtain

\frac{4}{2} = 2

. In the numerator, combine the bases by adding exponents:

x^{1/2} \cdot x^{1/3} = x^{1/2 + 1/3} = x^{5/6}

, and

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

. Next, divide by the terms in the denominator using the quotient rule:

\frac{x^{5/6}}{x^{-1}} = x^{5/6 - (-1)} = x^{11/6}

, and

\frac{y^2}{y^1} = y^{2 - 1} = y

. This yields the fully simplified expression

2x^{11/6}y

.

Step-by-Step Solution

1

Simplify the coefficients of the fraction.

\frac{4}{2} = 2

Dividing the numerical constants in the numerator and denominator simplifies the constant multiplier of the expression.

2

Combine the

x

terms in the numerator using the product rule.

x^{1/2} \cdot x^{1/3} = x^{1/2 + 1/3} = x^{5/6}

When multiplying terms with the same base, add their exponents:

\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

.

3

Combine the

y

terms in the numerator using the product rule.

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

When multiplying terms with the same base, add their exponents:

3 - 1 = 2

.

4

Simplify the

x

terms in the fraction using the quotient rule.

\frac{x^{5/6}}{x^{-1}} = x^{5/6 - (-1)} = x^{5/6 + 1} = x^{11/6}

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

5

Simplify the

y

terms in the fraction using the quotient rule.

\frac{y^2}{y^1} = y^{2 - 1} = y

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

6

Combine the simplified components.

2x^{11/6}y

Multiply the simplified coefficient,

x

term, and

y

term together to get the final simplified expression.

Key Concept

Properties of Exponents in Algebraic Expressions

Estimated Time:1m 30s

Question 25Question →

If the expression $\frac{(a^{-3} b^2)^{-2}}{a^4 (b^{-1})^3}$ is simplified to the form $a^x b^y$ , where $a$ and $b$ are positive real numbers, what is the value of the product $xy$ ?

-14

-2

2

10

18

Show answer & explanation

Answer: $-2$

Answer

-2

The correct answer is

-2

. Simplifying the expression step-by-step: first, apply the power of a product and power of a power rules to the numerator to get

a^6 b^{-4}

. Next, simplify the denominator to get

a^4 b^{-3}

. Then, apply the quotient rule to simplify the fraction to

a^{6-4} b^{-4-(-3)} = a^2 b^{-1}

. This gives

x=2

and

y=-1

. The product

xy

is

2 \cdot (-1) = -2

.

Step-by-Step Solution

1

Simplify the numerator using the power of a power rule

(x^p)^q = x^{pq}

and the power of a product rule

(xy)^p = x^p y^p

.

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

To raise a product to a power, raise each factor to that power by multiplying their exponents.

2

Simplify the denominator using the power of a power rule.

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

Multiply the exponents of

b

to simplify the term

(b^{-1})^3

.

3

Divide the simplified numerator by the simplified denominator using the quotient rule

\frac{x^p}{x^q} = x^{p-q}

.

\frac{a^6 b^{-4}}{a^4 b^{-3}} = a^{6-4} b^{-4 - (-3)} = a^2 b^{-1}

Subtract the exponent of the denominator from the exponent of the numerator for each base.

4

Identify the values of

x

and

y

from the simplified form

a^x b^y

and calculate the product

xy

.

x = 2

,

y = -1

, so

xy = 2 \cdot (-1) = -2

Match the simplified exponents with the variables

x

and

y

, then multiply them to find the final value.

Key Concept

Properties of Exponents in Algebraic Expressions

Alternative Method

Alternatively, you can rewrite the negative exponents as positive exponents by moving them across the fraction bar first. The numerator

(a^{-3} b^2)^{-2}

becomes

\frac{1}{(a^{-3} b^2)^2} = \frac{1}{a^{-6} b^4} = \frac{a^6}{b^4}

. The denominator

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

becomes

\frac{a^4}{b^3}

. Dividing these gives

\frac{a^6}{b^4} \cdot \frac{b^3}{a^4} = \frac{a^2}{b^1} = a^2 b^{-1}

. This yields

x = 2

and

y = -1

, leading to the product

xy = -2

.

Estimated Time:1m 30s

Question 26Question →

If $x$ and $y$ are non-zero real numbers such that the expression $\frac{(x^2 y^a)^3}{x^5 y^{-2}}$ is equivalent to $x^b y^{14}$ for some integers $a$ and $b$ , what is the value of $a + b$ ?

Show answer & explanation

Answer: 5

Answer

The value of

a + b

is

5

.

Simplifying the expression

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

using the exponent rules yields

x^{6-5} y^{3a-(-2)} = x^1 y^{3a+2}

. Equating this to

x^b y^{14}

shows that

b = 1

and

3a + 2 = 14

. Solving for

a

gives

a = 4

. Thus, the sum

a + b

is

4 + 1 = 5

.

Step-by-Step Solution

1

Simplify the numerator of the given expression.

x^6 y^{3a}

Applying the power of a product rule

(uv)^n = u^n v^n

and the power of a power rule

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

to

(x^2 y^a)^3

results in

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

.

2

Simplify the quotient by subtracting exponents with the same base.

x^1 y^{3a+2}

Using the quotient rule

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

, the base

x

term becomes

x^{6-5} = x^1

, and the base

y

term becomes

y^{3a - (-2)} = y^{3a+2}

.

3

Equate the exponents of like bases to find the values of

a

and

b

.

b = 1

and

a = 4

Comparing

x^1 y^{3a+2}

to

x^b y^{14}

gives

b = 1

and

3a + 2 = 14

. Solving

3a + 2 = 14

yields

3a = 12

, which simplifies to

a = 4

.

4

Sum the values of

a

and

b

.

5

Adding

a

and

b

yields

4 + 1 = 5

.

Key Concept

Applying properties of exponents, including the power of a product, power of a power, and quotient rules, to simplify algebraic expressions.

Estimated Time:1m 30s

Question 27Question →

For all real values of $x$ and $y$ where the expression is defined, simplify the algebraic fraction:

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

Which expression is equivalent to this fraction?

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

\frac{(x+y)^2}{y^3}

\frac{(x+y)^2}{y}

(x+y)^2 y^2

\frac{(x+y)^2 + 2}{y+2}

Show answer & explanation

Answer: $\frac{(x+y)^2}{y}$

Answer

The expression

\frac{(x+y)^2}{y}

The correct expression is derived by first simplifying the denominator

(xy^{-1})^{-2}

to

x^{-2}y^2

. We then group and simplify like terms:

\frac{(x+y)^3}{x+y} = (x+y)^2

,

\frac{x^{-2}}{x^{-2}} = 1

, and

\frac{y}{y^2} = \frac{1}{y}

. Multiplying these yields the simplified expression.

Step-by-Step Solution

1

Simplify the term

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

in the denominator

x^{-2} y^2

Apply the power of a product rule:

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

, which gives

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

. Then, use the power of a power rule:

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

.

2

Rewrite the original fraction with the simplified denominator

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

Substitute the simplified term back into the expression to align similar bases.

3

Simplify the fraction by dividing terms with like bases using the quotient rule

\frac{(x+y)^2}{y}

Divide each component:

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

,

\frac{x^{-2}}{x^{-2}} = x^{-2 - (-2)} = x^0 = 1

, and

\frac{y}{y^2} = y^{1-2} = y^{-1} = \frac{1}{y}

. Multiplying these results gives

\frac{(x+y)^2}{y}

.

Key Concept

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

Question 28Question →

If $w$ and $z$ are positive real numbers and $k$ is a constant such that the expression $\frac{(w^{3/2} z^{-1})^4}{(w z^k)^2}$ is equivalent to $w^4 z^6$ , what is the value of $k$ ?

Show answer & explanation

Answer: -5

Answer

The value of

k

is

-5

.

Applying the exponent rules, the expression simplifies to

w^4 z^{-4-2k}

. Equating the exponent of

z

to the exponent in the target expression

w^4 z^6

gives

-4-2k = 6

, which solves to

k = -5

.

Step-by-Step Solution

1

Apply the power of a product rule to the expression in the numerator

(w^{3/2} z^{-1})^4 = w^6 z^{-4}

When raising a product to a power, multiply the exponent of each factor by the outer exponent:

(x^a y^b)^c = x^{ac} y^{bc}

.

2

Apply the power of a product rule to the expression in the denominator

(w z^k)^2 = w^2 z^{2k}

Multiply the exponent of each factor in the denominator by

2

.

3

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

\frac{w^6 z^{-4}}{w^2 z^{2k}} = w^4 z^{-4-2k}

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

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

.

4

Equate the exponent of

z

in the simplified expression to the exponent of

z

in the target expression

-4 - 2k = 6

Since the simplified expression is equivalent to

w^4 z^6

, the exponents of the corresponding variable bases must be equal.

5

Solve the linear equation for

k

k = -5

Add

4

to both sides of the equation to get

-2k = 10

, then divide both sides by

-2

.

Key Concept

Properties of Exponents in Algebraic Expressions

Question 29Question →

For all positive real numbers $x$ and $y$ , consider the algebraic expression:

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

Which of the following is equivalent to this expression?

(x + y)^{5/6}

x^{5/3} + y^{5/3}

(x + y)^{5/3}

(x + y)^{11/10}

(x + y)^{5/3} y

Show answer & explanation

Answer: $(x + y)^{5/3}$

Answer

(x + y)^{5/3}

The correct answer is obtained by applying the properties of exponents step-by-step. First, apply the power of a product rule to the term

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

to get

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

. Next, combine the terms in the numerator using the product rule of exponents to get

(x+y)^{7/6}y

. Finally, divide by the denominator using the quotient rule of exponents, which cancels the

y

terms and results in

(x+y)^{5/3}

.

Step-by-Step Solution

1

Apply the power of a product property to simplify the first term in the numerator.

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

To distribute the exponent

1/2

to each factor inside the parentheses.

2

Multiply the simplified term by the other factor in the numerator.

(x + y)^{1/2}y \cdot (x + y)^{2/3} = (x + y)^{1/2 + 2/3}y = (x + y)^{7/6}y

To combine bases of the same value by adding their exponents:

\frac{1}{2} + \frac{2}{3} = \frac{3}{6} + \frac{4}{6} = \frac{7}{6}

.

3

Divide the simplified numerator by the denominator.

\frac{(x + y)^{7/6}y}{(x + y)^{-1/2}y} = (x + y)^{7/6 - (-1/2)} = (x + y)^{7/6 + 3/6} = (x + y)^{10/6} = (x + y)^{5/3}

To subtract the exponent in the denominator from the exponent in the numerator and cancel the common factor

y

.

Key Concept

Properties of Exponents in Algebraic Expressions

Alternative Method

Instead of simplifying the numerator first, we can divide the terms inside the parentheses first if we rewrite the expression, but the standard method of simplifying the numerator and then performing division is the most direct.

Estimated Time:1m 30s

Question 30Question →

For all non-zero real numbers $x$ and $y$ , the expression $\frac{(x^2 + y^{-1})^2}{x^4}$ is equivalent to which of the following?

1 + x^{-4}y^{-2}

1 + 2x^{-2}y + x^{-4}y^2

1 + 2x^{-2}y^{-1} + x^{-4}y^{-2}

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

y^{-2} + 2x^{-2}y^{-2} + x^{-4}y^{-2}

Show answer & explanation

Answer: $1 + 2x^{-2}y^{-1} + x^{-4}y^{-2}$

Answer

1 + 2x^{-2}y^{-1} + x^{-4}y^{-2}

The correct expression is obtained by expanding the binomial in the numerator using the perfect square formula

(a + b)^2 = a^2 + 2ab + b^2

. Substituting

a = x^2

and

b = y^{-1}

yields

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

. Dividing each term by the denominator

x^4

and applying the quotient rule for exponents

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

results in

1 + 2x^{-2}y^{-1} + x^{-4}y^{-2}

.

Step-by-Step Solution

1

Expand the binomial in the numerator.

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

Apply the binomial expansion formula

(a + b)^2 = a^2 + 2ab + b^2

where

a = x^2

and

b = y^{-1}

.

2

Divide each term in the expanded numerator by the denominator.

\frac{x^4}{x^4} + \frac{2x^2 y^{-1}}{x^4} + \frac{y^{-2}}{x^4}

Distribute the division over the terms in the numerator.

3

Simplify each fraction using exponent properties.

1 + 2x^{-2}y^{-1} + x^{-4}y^{-2}

Use the quotient rule

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

to simplify the variables.

Key Concept

Properties of Exponents in Algebraic Expressions

Question 31Question →

If $u$ and $v$ are positive real numbers such that $\frac{(u^{-2} v^3)^3}{u^k v^{-1}} = \frac{v^{10}}{u^{10}}$ , what is the value of the exponent $k$ ?

Show answer & explanation

Answer: 4

Answer

The value of the exponent is 4.

By applying the exponent rules systematically, the expression on the left simplifies to

u^{-6-k} v^{10}

, and the expression on the right is

u^{-10} v^{10}

. Equating the exponents of

u

gives

-6-k = -10

, which solves to

k = 4

.

Step-by-Step Solution

1

Apply the power of a product and power of a power properties to the numerator

(u^{-2} v^3)^3

.

u^{-6} v^9

According to the power of a product rule,

(xy)^a = x^a y^a

, and the power of a power rule,

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

.

2

Use the quotient of powers property to simplify the left side of the equation.

u^{-6-k} v^{10}

The quotient of powers rule states that

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

, so the exponents of like bases are subtracted:

-6 - k

for

u

and

9 - (-1) = 10

for

v

.

3

Rewrite the right side of the equation,

\frac{v^{10}}{u^{10}}

, using a negative exponent.

u^{-10} v^{10}

Applying the negative exponent rule,

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

.

4

Set the simplified expressions equal and solve for

k

.

k = 4

Since

u^{-6-k} v^{10} = u^{-10} v^{10}

, the exponents of the base

u

must be equal. Therefore,

-6 - k = -10

, which simplifies to

k = 4

.

Key Concept

Properties of Exponents in Algebraic Expressions

Estimated Time:1m 30s

Question 32Question →

Which of the following expressions is equivalent to $\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}}$ for all non-zero real numbers $a$ and $b$ where $a \neq b$ ?

a^2 - b^2

\frac{1}{a+b}

\frac{a+b}{ab}

\frac{1}{a-b}

a+b

Show answer & explanation

Answer: $\frac{a+b}{ab}$

Answer

The correct answer is

\frac{a+b}{ab}

because simplifying the original expression using the difference of squares or by rewriting negative exponents as fractions leads directly to this value.

By applying the difference of squares factorization to the numerator, the expression simplifies to

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

. Converting these to standard fractions and finding a common denominator yields

\frac{a+b}{ab}

.

Step-by-Step Solution

1

Factor the numerator as a difference of squares.

a^{-2} - b^{-2} = (a^{-1})^2 - (b^{-1})^2 = (a^{-1} - b^{-1})(a^{-1} + b^{-1})

Since the terms are squared with negative exponents, we can use the difference of squares identity

x^2 - y^2 = (x - y)(x + y)

where

x = a^{-1}

and

y = b^{-1}

.

2

Substitute the factored numerator back into the expression and cancel the common factor.

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

Since

a \neq b

, we know

a^{-1} - b^{-1} \neq 0

, allowing us to divide out the common factor in the numerator and denominator.

3

Rewrite the expression using positive exponents and find a common denominator.

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

A negative exponent

x^{-n}

represents the reciprocal

\frac{1}{x^n}

. Finding the common denominator

ab

allows us to combine the two fractions into a single expression.

Key Concept

Properties of negative exponents and difference of squares factorization

Alternative Method

Convert all terms to fractions with positive exponents first: write the numerator as

\frac{1}{a^2} - \frac{1}{b^2} = \frac{b^2 - a^2}{a^2b^2}

and the denominator as

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

. Then, divide the two complex fractions by multiplying the numerator by the reciprocal of the denominator:

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

. After canceling the common factors

(b-a)

and

ab

, the expression simplifies directly to

\frac{a+b}{ab}

.

Estimated Time:1m 30s

Question 33Question →

For all positive real numbers $w$ , the expression $\left(\frac{w^4 \cdot \sqrt{w}}{w^{-1.5}}\right)^{2/3}$ is equivalent to $w^k$ , where $k$ is a constant. What is the value of $k$ ?

Show answer & explanation

Answer: 4

Answer

The value of

k

is

4

.

Applying the exponent rules in sequence: first, rewrite the square root as a fractional exponent to get

w^{0.5}

. Next, multiply the terms in the numerator by adding their exponents:

4 + 0.5 = 4.5

. Then, divide the numerator by the denominator by subtracting the denominator's exponent from the numerator's exponent:

4.5 - (-1.5) = 6

. Finally, raise this result to the

2/3

power by multiplying the exponents:

6 \times (2/3) = 4

. This yields

w^4

, so the constant exponent is

4

.

Step-by-Step Solution

1

Convert the radical expression to an exponential expression.

\sqrt{w} = w^{0.5}

Converting all terms to base

w

with decimal or fractional exponents makes it easier to apply exponent properties.

2

Apply the product rule of exponents to the numerator.

w^4 \cdot w^{0.5} = w^{4 + 0.5} = w^{4.5}

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

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

.

3

Apply the quotient rule of exponents to the fraction.

\frac{w^{4.5}}{w^{-1.5}} = w^{4.5 - (-1.5)} = w^6

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

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

.

4

Apply the power rule of exponents to the simplified term.

(w^6)^{2/3} = w^{6 \cdot \frac{2}{3}} = w^4

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

(w^a)^b = w^{a \cdot b}

.

Key Concept

Properties of Exponents in Algebraic Expressions

Properties of Exponents in Algebraic Expressions

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes