Exponents, Radicals, and Algebraic Expressions Practice Questions — GMAT

Question 1Question →

For all positive real numbers $x$ and $y$ , if $x^2 y^3 = 108$ and $x^3 y^2 = 72$ , then $x + y = 5$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

Multiplying the two equations yields

(xy)^5 = 7776 = 6^5

, so

xy = 6

. Dividing the second equation by the first yields

\frac{x}{y} = \frac{2}{3}

, meaning

x = \frac{2}{3}y

. Substituting this into

xy = 6

gives

y = 3

and

x = 2

, making

x + y = 5

.

Step-by-Step Solution

1

Multiply the two system equations to find the product

xy

.

(x^2 y^3)(x^3 y^2) = x^5 y^5 = (xy)^5 = 108 \times 72 = 7776 = 6^5

, so

xy = 6

.

Multiplying exponential expressions with common bases allows adding their exponents to form a unified power

(xy)^5

.

2

Divide the second equation by the first equation to find the relationship between

x

and

y

.

\frac{x^3 y^2}{x^2 y^3} = \frac{x}{y} = \frac{72}{108} = \frac{2}{3}, which simplifies to

x = \frac{2}{3}y

.

Dividing exponential expressions subtracts their exponents, yielding the simple ratio of the variables.

3

Solve for individual values of

x

and

y

and evaluate

x + y

.

Substituting

x = \frac{2}{3}y

into

xy = 6

gives

\frac{2}{3}y^2 = 6 \implies y = 3

and

x = 2

. Therefore,

x + y = 5

.

Solving the system confirms that

x = 2

and

y = 3

are the unique positive real solutions.

Key Concept

System of exponential equations solved by combining power products and ratios.

Estimated Time:1m 30s

Question 2Question →

If $4^x + 4^x + 4^x + 4^x = 2^{10}$ , what is the value of $x$ ?

\frac{5}{4}

2

\frac{5}{2}

4

5

Show answer & explanation

Answer: $4$

Answer

The value of

x

is

4

.

Combining the four identical terms on the left side gives

4 \cdot 4^x = 4^{1+x}

. Converting

4

to base

2

yields

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

. Setting this equal to the right side

2^{10}

gives the linear equation

2x + 2 = 10

, which solves to

x = 4

.

Step-by-Step Solution

1

Combine the repeated addition on the left side of the equation.

4^x + 4^x + 4^x + 4^x = 4 \cdot 4^x = 4^{x+1}

Adding four identical terms is equivalent to multiplying the term by 4.

2

Convert all bases to 2 so that both sides can be compared.

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

Since

4 = 2^2

, applying the power of a power rule

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

expresses the left side with base 2.

3

Equate the exponents and solve for

x

.

2x + 2 = 10 \implies 2x = 8 \implies x = 4

If

b^u = b^v

for a positive base

b \neq 1

, then

u = v

.

Key Concept

Combining like exponential terms by factoring and applying base conversion rules

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

and

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

.

Question 3Question →

What is the numerical value of the expression $\sqrt[3]{7 + 5\sqrt{2}} - \sqrt[3]{5\sqrt{2} - 7}$ ?

Show answer & explanation

Answer: 2

Answer

The numerical value of the expression is 2.

The value of the expression is 2. This can be demonstrated either by setting the expression equal to

x

, cubing both sides to construct the cubic equation

x^3 + 3x - 14 = 0

, and factoring out the real root

x = 2

, or by recognizing that

(1 + \sqrt{2})^3 = 7 + 5\sqrt{2}

and

(\sqrt{2} - 1)^3 = 5\sqrt{2} - 7

, which simplifies the expression directly to

(1 + \sqrt{2}) - (\sqrt{2} - 1) = 2

.

Step-by-Step Solution

1

Define variables for the two cubic terms and write the target expression as a difference.

Let

u = \sqrt[3]{7 + 5\sqrt{2}}

and

v = \sqrt[3]{5\sqrt{2} - 7}

, so the target value is

x = u - v

.

Grouping nested radical terms simplifies the algebraic manipulation.

2

Cube both sides of

x = u - v

using the algebraic identity

(u - v)^3 = u^3 - v^3 - 3uv(u - v)

.

x^3 = u^3 - v^3 - 3uv \cdot x

.

Cubing eliminates the outer radical signs on the cubed terms.

3

Evaluate

u^3 - v^3

and the product

uv

.

u^3 - v^3 = (7 + 5\sqrt{2}) - (5\sqrt{2} - 7) = 14

, and

uv = \sqrt[3]{(5\sqrt{2}+7)(5\sqrt{2}-7)} = \sqrt[3]{50 - 49} = 1

.

Using the difference of squares under the cube root simplifies the product term to 1.

4

Substitute the evaluated terms into the cubic equation and solve for the real root

x

.

x^3 = 14 - 3x \implies x^3 + 3x - 14 = 0 \implies (x - 2)(x^2 + 2x + 7) = 0 \implies x = 2

.

The quadratic factor

x^2 + 2x + 7

has negative discriminant (

4 - 28 = -24

), leaving

x = 2

as the unique real solution.

Key Concept

Simplifying nested radicals using cubic algebraic identities and binomial expansions

Question 4Question →

If $x$ and $y$ are positive integers such that $2^x - 2^{x-2} = 3^y$ , what is the value of $2^{x+y} + 3^{x-y}$ ?

11

13

25

35

73

Show answer & explanation

Answer: 11

Answer

The value of

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

is 11.

Factoring the left side of the equation yields

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

. Setting this equal to

3^y

and dividing both sides by 3 gives

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

. Because 2 and 3 are coprime bases, equality is possible only when both exponents equal zero. This determines

x = 2

and

y = 1

, both of which are positive integers. Substituting these values into

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

results in

2^3 + 3^1 = 8 + 3 = 11

.

Step-by-Step Solution

1

Factor out the common term on the left side of the given equation

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

Factoring out the smallest power of 2 simplifies the subtraction into a multiplication.

2

Equate the factored expression to the right side and simplify

3 \cdot 2^{x-2} = 3^y \implies 2^{x-2} = \frac{3^y}{3} = 3^{y-1}

Dividing both sides by 3 isolates the prime bases 2 and 3 on opposite sides of the equation.

3

Apply fundamental properties of prime factorizations

x - 2 = 0 \implies x = 2

and

y - 1 = 0 \implies y = 1

Since 2 and 3 are distinct prime numbers,

2^a = 3^b

can hold for integer exponents only if

a = 0

and

b = 0

.

4

Evaluate the target expression using

x = 2

and

y = 1

2^{2+1} + 3^{2-1} = 2^3 + 3^1 = 8 + 3 = 11

Substituting the unique positive integer solution pair

(2, 1)

into the expression yields the final value.

Key Concept

Uniqueness of Prime Base Exponent Equality

Question 5Question →

If $a$ and $b$ are positive integers such that $2^{a+3} \cdot 5^b - 2^a \cdot 5^{b+1} = 6,000$ , what is the value of $a \cdot b$ ?

7

10

12

15

16

Show answer & explanation

Answer: 12

Answer

12

Factoring out

2^a \cdot 5^b

yields

2^a \cdot 5^b (2^3 - 5^1) = 3 \cdot 2^a \cdot 5^b = 6,000

. Dividing by 3 yields

2^a \cdot 5^b = 2,000

. Expressing

2,000

in prime factored form gives

2^4 \cdot 5^3

. Matching exponents for the prime bases gives

a = 4

and

b = 3

. The product

a \cdot b

is

4 \times 3 = 12

.

Step-by-Step Solution

1

Rewrite the terms in the expression using exponent rules to isolate common bases.

2^{a+3} \cdot 5^b - 2^a \cdot 5^{b+1} = (2^a \cdot 2^3) \cdot 5^b - 2^a \cdot (5^b \cdot 5^1)

Applying

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

allows us to extract common powers of

2^a

and

5^b

.

2

Factor out the common term

2^a \cdot 5^b

from the left side of the equation.

2^a \cdot 5^b (2^3 - 5^1) = 2^a \cdot 5^b (8 - 5) = 3 \cdot 2^a \cdot 5^b

Simplifying the constant factor in parentheses simplifies the equation.

3

Divide both sides of the equation by 3 and perform prime factorization on the resulting integer.

3 \cdot 2^a \cdot 5^b = 6,000 \implies 2^a \cdot 5^b = 2,000 = 2^4 \cdot 5^3

Prime factorization of

2,000

determines the unique integer exponents for bases 2 and 5.

4

Equate the corresponding exponents and calculate the requested product

a \cdot b

.

a = 4

and

b = 3 \implies a \cdot b = 4 \times 3 = 12

Since 2 and 5 are prime numbers, the prime factorization representation is unique.

Key Concept

Factoring Exponents and Prime Factorization

Estimated Time:2m 0s

Question 6Question →

If $x = \sqrt[3]{5 + 2\sqrt{13}} + \sqrt[3]{5 - 2\sqrt{13}}$ , what is the value of $x^3 + 9x$ ?

Show answer & explanation

Answer: 10

Answer

10

By defining

x = u + v

with

u = \sqrt[3]{5 + 2\sqrt{13}}

and

v = \sqrt[3]{5 - 2\sqrt{13}}

, cubing both sides gives

x^3 = u^3 + v^3 + 3uv(u + v)

. Evaluating the components yields

u^3 + v^3 = 10

and

uv = \sqrt[3]{25 - 52} = -3

. Substituting these back gives

x^3 = 10 - 9(x)

, which rearranges to

x^3 + 9x = 10

.

Step-by-Step Solution

1

Express

x

as a sum of two variables

u

and

v

x = u + v

, where

u = \sqrt[3]{5 + 2\sqrt{13}}

and

v = \sqrt[3]{5 - 2\sqrt{13}}

Grouping the binomial terms simplifies algebraic expansion using standard polynomial identities.

2

Calculate the sum of the cubes

u^3 + v^3

u^3 + v^3 = (5 + 2\sqrt{13}) + (5 - 2\sqrt{13}) = 10

Eliminating the cube roots allows for simple additive cancellation of the radical terms.

3

Calculate the product

uv

uv = \sqrt[3]{(5 + 2\sqrt{13})(5 - 2\sqrt{13})} = \sqrt[3]{5^2 - (2\sqrt{13})^2} = \sqrt[3]{25 - 52} = \sqrt[3]{-27} = -3

Applying the difference of squares property inside the cube root simplifies the product of conjugate radicals to a single integer.

4

Cube both sides of

x = u + v

and substitute evaluated terms

x^3 = u^3 + v^3 + 3uv(u + v) = 10 + 3(-3)x = 10 - 9x

Using

(u+v)^3 = u^3 + v^3 + 3uv(u+v)

connects

x^3

directly to

x

without expanding long radical terms.

5

Rearrange the equation to isolate

x^3 + 9x

x^3 + 9x = 10

Adding

9x

to both sides gives the exact numerical value of the requested expression.

Key Concept

Algebraic Identities with Polynomials and Radicals

Question 7Question →

If $x = \sqrt{7 - 4\sqrt{3}}$ and $y = \sqrt{7 + 4\sqrt{3}}$ , what is the value of $\frac{x^{-2} + y^{-2}}{(x+y)^{-1}}$ ?

\frac{1}{4}

\frac{7}{2}

14

56

64

Show answer & explanation

Answer: $56$

Answer

56

The expression

x = \sqrt{7 - 4\sqrt{3}}

can be unnested by recognizing

7 - 4\sqrt{3}

as the perfect square

(2 - \sqrt{3})^2

, so

x = 2 - \sqrt{3}

and

y = 2 + \sqrt{3}

. This gives

x+y = 4

and

xy = 1

. The numerator

x^{-2} + y^{-2} = \frac{x^2+y^2}{(xy)^2} = \frac{(x+y)^2 - 2xy}{1} = 14

. The denominator

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

. Dividing

14

by

\frac{1}{4}

yields

56

.

Step-by-Step Solution

1

Simplify the radical expressions for

x

and

y

.

x = 2 - \sqrt{3}

and

y = 2 + \sqrt{3}

.

Note that

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

, so

\sqrt{7 - 4\sqrt{3}} = 2 - \sqrt{3}

. Similarly,

7 + 4\sqrt{3} = (2 + \sqrt{3})^2

.

2

Find the sum

x+y

and product

xy

.

x+y = 4

and

xy = 1

.

(2 - \sqrt{3}) + (2 + \sqrt{3}) = 4

, and

(2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1

.

3

Simplify the numerator

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

.

x^{-2} + y^{-2} = 14

.

x^{-2} + y^{-2} = \frac{1}{x^2} + \frac{1}{y^2} = \frac{x^2 + y^2}{(xy)^2} = \frac{(x+y)^2 - 2xy}{(xy)^2} = \frac{4^2 - 2(1)}{1^2} = 14

.

4

Evaluate the full expression

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

.

\frac{14}{1/4} = 56

.

The denominator is

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

. Dividing

14

by

\frac{1}{4}

equals

14 \cdot 4 = 56

.

Key Concept

Nested Radical Simplification & Algebraic Exponent Identities

Alternative Method

Instead of unnesting the radicals first, observe that

x^2 = 7 - 4\sqrt{3}

and

y^2 = 7 + 4\sqrt{3}

. Then

x^2 y^2 = (7-4\sqrt{3})(7+4\sqrt{3}) = 49 - 48 = 1

, and

x^2 + y^2 = 14

. Thus

x^{-2} + y^{-2} = \frac{x^2+y^2}{x^2 y^2} = 14

. Next, find

(x+y)^2 = x^2 + y^2 + 2xy = 14 + 2(1) = 16

, so

x+y = 4

. Then

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

, leading directly to

\frac{14}{1/4} = 56

.

Estimated Time:2m 0s

Question 8Question →

If $x = \sqrt{7 + 4\sqrt{3}}$ , what is the value of $x - \frac{1}{x}$ ?

2

4

\sqrt{3}

2\sqrt{3}

4\sqrt{3}

Show answer & explanation

Answer: $2\sqrt{3}$

Answer

2\sqrt{3}

The expression under the outer radical,

7 + 4\sqrt{3}

, can be rewritten as

(2 + \sqrt{3})^2

because

2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}

. Taking the square root gives

x = 2 + \sqrt{3}

. The reciprocal

\frac{1}{x}

is

\frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}

. Thus,

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

.

Step-by-Step Solution

1

Un-nest the radical

\sqrt{7 + 4\sqrt{3}}

by expressing

7 + 4\sqrt{3}

as a perfect square

(a + b\sqrt{3})^2

.

(a + b\sqrt{3})^2 = a^2 + 3b^2 + 2ab\sqrt{3} = 7 + 4\sqrt{3}

, which yields

a = 2

and

b = 1

, so

x = 2 + \sqrt{3}

.

Recognizing nested radicals in the form

\sqrt{A + B\sqrt{C}}

allows simplification into a binomial radical.

2

Find the reciprocal

\frac{1}{x}

by rationalizing the denominator.

\frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.

Multiplying the numerator and denominator by the conjugate clears the radical from the denominator.

3

Calculate the difference

x - \frac{1}{x}

.

(2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2\sqrt{3}.

Subtracting the reciprocal isolates the irrational component.

Key Concept

Simplification of nested radicals and rationalizing denominators using algebraic conjugates.

Estimated Time:1m 30s

Question 9Question →

If $3^x + 3^{x+1} + 3^{x+2} = 351$ , what is the value of $(x + 1)^x$ ?

12

27

64

81

243

Show answer & explanation

Answer: $64$

Answer

The value of

(x + 1)^x

is

64

.

Factoring

3^x

out of the sum gives

3^x(1 + 3 + 9) = 13 \cdot 3^x = 351

. Dividing

351

by

13

gives

3^x = 27

, so

x = 3

. Substituting

x = 3

into

(x + 1)^x

yields

(3 + 1)^3 = 4^3 = 64

.

Step-by-Step Solution

1

Factor out the common exponential term

3^x

from the left side of the equation.

3^x(1 + 3^1 + 3^2) = 351

, which simplifies to

3^x(1 + 3 + 9) = 351

or

13 \cdot 3^x = 351

.

When adding terms with identical bases and variable exponents, factor out the term with the smallest exponent.

2

Solve for

3^x

and determine the value of

x

.

3^x = \frac{351}{13} = 27

. Since

27 = 3^3

,

x = 3

.

Divide both sides by

13

and set the powers of equal bases equal to find

x

.

3

Substitute

x = 3

into the target expression

(x + 1)^x

.

(3 + 1)^3 = 4^3 = 64

.

Evaluate the expression as requested by the question stem.

Key Concept

Factoring exponential expressions with common bases and applying exponent rules

Estimated Time:1m 30s

Question 10Question →

For all real numbers $x$ , the identity $\sqrt{x^2} = x$ holds true.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false because the principal square root

\sqrt{x^2}

is defined as the absolute value of

x

,

|x|

, which equals

-x

when

x < 0

.

The statement is false because the expression

\sqrt{x^2}

represents the principal (non-negative) square root of

x^2

, which simplifies to

|x|

. When

x

is negative,

|x| = -x

, which is not equal to

x

.

Step-by-Step Solution

1

Recall the definition of the principal square root radical symbol

\sqrt{\vphantom{x}}

By definition,

\sqrt{a} \ge 0

for any real number

a \ge 0

.

The principal square root function always yields a non-negative output.

2

Test a negative value for the variable

x

Let

x = -3

. Then

\sqrt{(-3)^2} = \sqrt{9} = 3

.

Evaluating a specific negative number tests if the equality holds universally for all real numbers.

3

Compare the evaluated result with the original value of

x

Since

3 \neq -3

,

\sqrt{x^2} \neq x

when

x < 0

.

A single counterexample disproves a universal mathematical identity statement.

Key Concept

Principal Square Root and Absolute Value Property (\sqrt{x^2} = |x|)

Estimated Time:45s

Question 11Question →

For all real numbers $x > 1$ , the expression $\sqrt{x + 2\sqrt{x - 1}} - \sqrt{x - 2\sqrt{x - 1}}$ is equal to $2$ .

True

False

Show answer & explanation

Answer: False

Answer

The statement is False because the expression equals

2\sqrt{x-1}

(which is less than

2

) whenever

1 < x < 2

.

The statement asserts that the identity holds for all

x > 1

. However, evaluating

\sqrt{x - 2\sqrt{x-1}}

yields

|\sqrt{x-1} - 1|

. When

1 < x < 2

,

\sqrt{x-1} < 1

, so

|\sqrt{x-1} - 1| = 1 - \sqrt{x-1}

. Subtraction gives

2\sqrt{x-1} \neq 2

for all

x

in

(1, 2)

. Thus, the statement is false.

Step-by-Step Solution

1

Rewrite the expressions under the outer radicals as perfect squares.

x \pm 2\sqrt{x-1} = (x-1) \pm 2\sqrt{x-1} + 1 = (\sqrt{x-1} \pm 1)^2

Recognize the quadratic pattern

a^2 \pm 2ab + b^2

where

a = \sqrt{x-1}

and

b = 1

.

2

Apply the radical identity

\sqrt{u^2} = |u|

to remove the outer square roots.

\sqrt{x + 2\sqrt{x-1}} = |\sqrt{x-1} + 1| = \sqrt{x-1} + 1

and

\sqrt{x - 2\sqrt{x-1}} = |\sqrt{x-1} - 1|

The principal square root of a squared quantity is non-negative, requiring absolute value bars.

3

Analyze the absolute value

|\sqrt{x-1} - 1|

across the specified domain

x > 1

.

For

x \ge 2

,

\sqrt{x-1} \ge 1 \implies |\sqrt{x-1} - 1| = \sqrt{x-1} - 1

. For

1 < x < 2

,

\sqrt{x-1} < 1 \implies |\sqrt{x-1} - 1| = 1 - \sqrt{x-1}

.

The sign of the expression inside the absolute value changes at

x = 2

.

4

Evaluate the full expression for

1 < x < 2

.

(\sqrt{x-1} + 1) - (1 - \sqrt{x-1}) = 2\sqrt{x-1}

Since

2\sqrt{x-1} < 2

when

1 < x < 2

, the statement does not hold for all real numbers

x > 1

.

Key Concept

Nested Radicals and the Principal Square Root Absolute Value Identity

\sqrt{u^2} = |u|

Question 12Question →

If $x = \sqrt{3} + \sqrt{2}$ , what is the value of $x^4 + x^{-4}$ ?

98

100

102

144

14

Show answer & explanation

Answer: 98

Answer

98

Rationalizing the denominator gives

x^{-1} = \sqrt{3} - \sqrt{2}

. Summing

x + x^{-1}

yields

2\sqrt{3}

. Squaring this sum gives

x^2 + 2 + x^{-2} = 12

, so

x^2 + x^{-2} = 10

. Squaring once more gives

x^4 + 2 + x^{-4} = 100

, which simplifies to

x^4 + x^{-4} = 98

.

Step-by-Step Solution

1

Find the reciprocal expression

x^{-1}

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

Rationalizing the denominator simplifies the expression for

x^{-1}

.

2

Calculate the sum

x + x^{-1}

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

Adding the original expression and its reciprocal eliminates the

\sqrt{2}

term.

3

Square

x + x^{-1}

to find

x^2 + x^{-2}

(x + x^{-1})^2 = x^2 + 2(x)(x^{-1}) + x^{-2} = x^2 + 2 + x^{-2} = (2\sqrt{3})^2 = 12 \implies x^2 + x^{-2} = 10

Using the algebraic identity

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

allows finding

x^2 + x^{-2}

.

4

Square

x^2 + x^{-2}

to find

x^4 + x^{-4}

(x^2 + x^{-2})^2 = x^4 + 2(x^2)(x^{-2}) + x^{-4} = x^4 + 2 + x^{-4} = 10^2 = 100 \implies x^4 + x^{-4} = 98

Squaring the second-power sum and subtracting the cross-term 2 yields the required value.

Key Concept

Rationalizing radical denominators and binomial squaring of reciprocals

Estimated Time:1m 30s

Question 13Question →

For all positive real numbers $x$ , the nested radical expression $\sqrt{x\sqrt{x}}$ is mathematically equivalent to $x^{\frac{3}{4}}$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because expressing

\sqrt{x\sqrt{x}}

in fractional exponent form yields

x^{\frac{3}{4}}

for all positive real numbers

x

.

The radical expression

\sqrt{x\sqrt{x}}

simplifies to

x^{\frac{3}{4}}

by systematically applying exponent laws: the inner root gives

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

, multiplying by

x

gives

x^{\frac{3}{2}}

, and applying the outer root raises

x^{\frac{3}{2}}

to the

\frac{1}{2}

power, producing

x^{\frac{3}{4}}

.

Step-by-Step Solution

1

Express the inner radical as a fractional exponent.

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

, making the expression under the outer root

x \cdot x^{\frac{1}{2}}

.

By definition of fractional exponents,

\sqrt[n]{a^m} = a^{\frac{m}{n}}

.

2

Combine the terms inside the outer radical using the product rule for exponents.

x^1 \cdot x^{\frac{1}{2}} = x^{1 + \frac{1}{2}} = x^{\frac{3}{2}}

.

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

3

Apply the outer square root as an exponent of

\frac{1}{2}

and simplify using the power rule.

\sqrt{x^{\frac{3}{2}}} = (x^{\frac{3}{2}})^{\frac{1}{2}} = x^{\frac{3}{2} \cdot \frac{1}{2}} = x^{\frac{3}{4}}

.

Raising a power to another power requires multiplying the exponents.

Key Concept

Simplification of nested radical expressions using fractional exponent laws

Question 14Question →

If $y = 3^k$ , which of the following expressions is equivalent to $9^{k+1}$ for all real values of $k$ ?

9y^2

9y

3y^2

y^2 + 1

y^3

Show answer & explanation

Answer: $9y^2$

Answer

The expression

9y^2

is equivalent to

9^{k+1}

.

Applying the product rule of exponents gives

9^{k+1} = 9^k \cdot 9^1

. Expressing 9 as

3^2

gives

9^k = (3^2)^k = (3^k)^2

. Substituting

y = 3^k

yields

9 \cdot y^2 = 9y^2

.

Step-by-Step Solution

1

Rewrite

9^{k+1}

using the product rule of exponents (

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

).

9^{k+1} = 9^k \cdot 9^1 = 9 \cdot 9^k

Separating the exponent addition into a product allows simplification of the constant factor.

2

Express the base 9 in terms of base 3.

9^k = (3^2)^k = (3^k)^2

Since

y = 3^k

, converting base 9 to

3^2

allows substitution of

y

into the expression.

3

Substitute

y = 3^k

into the expression.

9 \cdot (3^k)^2 = 9y^2

Replacing

3^k

with

y

yields the simplified algebraic expression in terms of

y

.

Key Concept

Properties of exponents including power of a power

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

and product of powers

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

.

Estimated Time:45s

Question 15Question →

What is the numerical value of $\frac{2^8 \cdot 3^4}{6^4}$ ?

Show answer & explanation

Answer: 16

Answer

16

Rewriting the base

6

as

2 \cdot 3

gives

6^4 = 2^4 \cdot 3^4

. Substituting this into the original expression yields

\frac{2^8 \cdot 3^4}{2^4 \cdot 3^4}

. Cancelling

3^4

leaves

\frac{2^8}{2^4} = 2^{8-4} = 2^4 = 16

.

Step-by-Step Solution

1

Rewrite composite bases into prime factors

6^4 = (2 \cdot 3)^4 = 2^4 \cdot 3^4

Applying the power of a product rule

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

allows base matching.

2

Simplify the quotient using exponent subtraction

\frac{2^8 \cdot 3^4}{2^4 \cdot 3^4} = 2^{8-4} \cdot 3^{4-4} = 2^4 \cdot 3^0 = 16 \cdot 1 = 16

Applying the quotient rule

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

for powers with equal bases.

Key Concept

Exponent rules: Power of a Product

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

and Quotient of Powers

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

Question 16Question →

What is the value of the expression $\frac{3^8 + 3^8 + 3^8}{3^4}$ ?

3^4

3^5

3^8

3^{12}

3^{20}

Show answer & explanation

Answer: $3^5$

Answer

3^5

The correct answer is

3^5

. Adding three identical values

3^8 + 3^8 + 3^8

is equivalent to

3 \times 3^8

, which equals

3^9

. Dividing

3^9

by

3^4

gives

3^{9-4} = 3^5

.

Step-by-Step Solution

1

Simplify the numerator by combining like terms

3^8 + 3^8 + 3^8 = 3 \cdot 3^8 = 3^1 \cdot 3^8 = 3^{1+8} = 3^9

Repeated addition of

3

identical terms is equivalent to multiplying one term by

3

.

2

Apply the quotient rule for exponents

\frac{3^9}{3^4} = 3^{9-4} = 3^5

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

Key Concept

Combining like exponential terms using multiplication rules and applying the quotient rule for exponents

Question 17Question →

If $x$ is a real number, what is the sum of all valid real solutions to the equation $\sqrt{3x - 2} - \sqrt{x + 3} = 1$ ?

1

3

4

6

7

Show answer & explanation

Answer: 6

Answer

The sum of all valid real solutions is 6.

To solve the radical equation, we isolate

\sqrt{3x - 2}

to get

\sqrt{3x - 2} = 1 + \sqrt{x + 3}

. Squaring both sides yields

3x - 2 = x + 4 + 2\sqrt{x + 3}

, which simplifies to

x - 3 = \sqrt{x + 3}

. Squaring both sides a second time produces the quadratic

x^2 - 7x + 6 = 0

, giving candidate solutions

x = 6

and

x = 1

. Checking in the original equation reveals that

x = 6

yields

\sqrt{16} - \sqrt{9} = 1

, which is true, whereas

x = 1

yields

\sqrt{1} - \sqrt{4} = -1 \neq 1

, making

x = 1

extraneous. The only valid solution is 6, so the sum of all valid solutions is 6.

Step-by-Step Solution

1

Isolate one radical expression on one side of the equation.

\sqrt{3x - 2} = 1 + \sqrt{x + 3}

Isolating a radical allows for clean squaring of both sides without creating cross-terms with multiple roots.

2

Square both sides of the equation.

3x - 2 = 1 + 2\sqrt{x + 3} + (x + 3) \implies 3x - 2 = x + 4 + 2\sqrt{x + 3}

Eliminates the radical on the left side.

3

Isolate the remaining radical term and simplify.

2x - 6 = 2\sqrt{x + 3} \implies x - 3 = \sqrt{x + 3}

Simplifies the algebraic expression prior to squaring a second time.

4

Square both sides again to obtain a quadratic equation.

(x - 3)^2 = x + 3 \implies x^2 - 6x + 9 = x + 3 \implies x^2 - 7x + 6 = 0

Eliminates all remaining radical terms.

5

Factor the quadratic equation to find algebraic candidate solutions.

(x - 6)(x - 1) = 0 \implies x = 6 \text{ or } x = 1

Identifies potential real roots.

6

Test candidate solutions in the original equation

\sqrt{3x - 2} - \sqrt{x + 3} = 1

to remove extraneous solutions.

For

x = 6

:

\sqrt{16} - \sqrt{9} = 4 - 3 = 1

(Valid). For

x = 1

:

\sqrt{1} - \sqrt{4} = 1 - 2 = -1 \neq 1

(Extraneous). Thus,

x = 6

is the sole valid solution.

Squaring equations during algebraic steps can introduce extraneous roots that do not satisfy the original radical equation.

Key Concept

Solving Radical Equations and Eliminating Extraneous Solutions

Estimated Time:2m 0s

Question 18Question →

If $5^x = 100^y = 20^z$ for non-zero real numbers $x$ , $y$ , and $z$ , which of the following expressions is equal to $z$ ?

\frac{xy}{x - y}

\frac{xy}{x + y}

\frac{xy}{y - x}

x - y

y - x

Show answer & explanation

Answer: $\frac{xy}{x - y}$

Answer

\frac{xy}{x - y}

By setting

5^x = 100^y = 20^z = K

, we rewrite each base as a power of

K

:

5 = K^{1/x}

,

100 = K^{1/y}

, and

20 = K^{1/z}

. Using the numerical relationship

20 = \frac{100}{5}

, we substitute to obtain

K^{1/z} = \frac{K^{1/y}}{K^{1/x}} = K^{\frac{1}{y} - \frac{1}{x}}

. Equating exponents gives

\frac{1}{z} = \frac{1}{y} - \frac{1}{x} = \frac{x - y}{xy}

. Taking the reciprocal yields the expression stating that

z

equals

\frac{xy}{x - y}

.

Step-by-Step Solution

1

Set the common exponential expressions equal to a constant

K

.

Let

5^x = 100^y = 20^z = K

, where

K > 0

and

K \neq 1

.

Equating exponential terms to a shared parameter allows conversion from variable powers to fractional exponents.

2

Express the base numbers

5

,

100

, and

20

in terms of

K

.

5 = K^{1/x}

,

100 = K^{1/y}

, and

20 = K^{1/z}

.

Taking the

x

-th,

y

-th, and

z

-th roots of both sides isolates each base number.

3

Relate the base numbers using arithmetic division.

20 = \frac{100}{5} \implies K^{1/z} = \frac{K^{1/y}}{K^{1/x}} = K^{\frac{1}{y} - \frac{1}{x}}

.

Dividing powers with the same base subtracts their exponents.

4

Equate the exponents and solve for

z

.

\frac{1}{z} = \frac{1}{y} - \frac{1}{x} = \frac{x - y}{xy} \implies z = \frac{xy}{x - y}

.

Taking common denominators and inverting the fraction yields

z

in terms of

x

and

y

.

Key Concept

Relating different bases with variable exponents by setting them equal to a common constant and converting to reciprocal exponents.

Estimated Time:2m 0s

Question 19Question →

For all real numbers $x > 0$ such that $x \neq 1$ , the algebraic expression $\frac{\sqrt{x} - 1}{\sqrt[4]{x} + 1} + \frac{\sqrt{x} + 1}{\sqrt[4]{x} - 1}$ is equivalent to $\frac{2(\sqrt[4]{x}^3 + 1)}{\sqrt{x} - 1}$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

The statement is true because substituting

u = \sqrt[4]{x}

allows both expressions to be simplified via algebraic factoring identities to the identical expression

\frac{2(u^2 - u + 1)}{u - 1}

.

Step-by-Step Solution

1

Perform a substitution to simplify the radical exponents.

Let

u = \sqrt[4]{x}

, so that

\sqrt{x} = u^2

. The left-hand side becomes

\frac{u^2 - 1}{u + 1} + \frac{u^2 + 1}{u - 1}

.

Converting fourth roots and square roots into polynomial terms makes factoring easier.

2

Simplify the left-hand side expression.

\frac{(u-1)(u+1)}{u+1} + \frac{u^2+1}{u-1} = (u-1) + \frac{u^2+1}{u-1} = \frac{(u-1)^2 + u^2 + 1}{u-1} = \frac{2(u^2 - u + 1)}{u-1}.

Factoring the numerator of the first fraction cancels out the

(u+1)

term prior to combining terms.

3

Factor the right-hand side expression using polynomial identities.

\frac{2(u^3+1)}{u^2-1} = \frac{2(u+1)(u^2-u+1)}{(u+1)(u-1)} = \frac{2(u^2-u+1)}{u-1}.

Applying the sum of cubes identity

u^3 + 1 = (u + 1)(u^2 - u + 1)

allows cancellation of

(u+1)

from the denominator.

4

Compare the simplified left-hand side and right-hand side expressions.

Both expressions reduce to

\frac{2(\sqrt[4]{x}^2 - \sqrt[4]{x} + 1)}{\sqrt[4]{x} - 1}

.

Because both sides reduce to the identical simplified form for all

x > 0, x \neq 1

, the equivalence holds true.

Key Concept

Algebraic manipulation of radicals using fractional exponent substitution, difference of squares, and sum of cubes factoring.

Question 20Question →

If $x$ is a positive real number such that $x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 4$ , what is the value of $x^{\frac{3}{2}} + x^{-\frac{3}{2}}$ ?

48

52

56

61

64

Show answer & explanation

Answer: 52

Answer

52

Cubing both sides of

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

yields

(x^{\frac{1}{2}} + x^{-\frac{1}{2}})^3 = 64

. By the identity

(a+b)^3 = a^3 + b^3 + 3ab(a+b)

, the left side expands to

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

. Substituting

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

and

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

gives

(x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 12 = 64

. Subtracting 12 yields 52.

Step-by-Step Solution

1

Set up the cubic identity for the sum of fractional exponents.

Let

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

and

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

. Then

a + b = 4

and

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

.

Recognizing that

x^{\frac{3}{2}} = a^3

and

x^{-\frac{3}{2}} = b^3

allows the use of binomial expansion.

2

Cube both sides of the given equation

a + b = 4

.

(a + b)^3 = a^3 + b^3 + 3ab(a + b) \implies 4^3 = (x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 3(1)(4)

.

Expanding the cube retains the desired expression

a^3 + b^3

alongside simpler terms.

3

Simplify and solve for

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

.

64 = (x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 12 \implies x^{\frac{3}{2}} + x^{-\frac{3}{2}} = 64 - 12 = 52.

Subtracting 12 from 64 gives the exact value of the expression.

Key Concept

Algebraic identities with fractional exponents

Estimated Time:2m 0s

Exponents, Radicals, and Algebraic Expressions

Answer

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