Exponents, Radicals, and Algebraic Expressions Alıştırma Soruları — GMAT

Soru 1Soru →

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

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Tahmini Süre:1m 30s

Soru 2Soru →

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

Cevabı ve açıklamayı göster

Cevap: $4$

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

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}

.

Soru 3Soru →

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

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Simplifying nested radicals using cubic algebraic identities and binomial expansions

Soru 4Soru →

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

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Uniqueness of Prime Base Exponent Equality

Soru 5Soru →

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

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Factoring Exponents and Prime Factorization

Tahmini Süre:2m 0s

Soru 6Soru →

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

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Algebraic Identities with Polynomials and Radicals

Soru 7Soru →

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

Cevabı ve açıklamayı göster

Cevap: $56$

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Nested Radical Simplification & Algebraic Exponent Identities

Alternatif Yöntem

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

.

Tahmini Süre:2m 0s

Soru 8Soru →

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}

Cevabı ve açıklamayı göster

Cevap: $2\sqrt{3}$

Cevap

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}

.

Adım Adım Çözüm

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.

Anahtar Kavram

Simplification of nested radicals and rationalizing denominators using algebraic conjugates.

Tahmini Süre:1m 30s

Soru 9Soru →

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

12

27

64

81

243

Cevabı ve açıklamayı göster

Cevap: $64$

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Factoring exponential expressions with common bases and applying exponent rules

Tahmini Süre:1m 30s

Soru 10Soru →

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

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Tahmini Süre:45s

Soru 11Soru →

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

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

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.

Adım Adım Çözüm

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

.

Anahtar Kavram

Nested Radicals and the Principal Square Root Absolute Value Identity

\sqrt{u^2} = |u|

Soru 12Soru →

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

98

100

102

144

14

Cevabı ve açıklamayı göster

Cevap: 98

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Rationalizing radical denominators and binomial squaring of reciprocals

Tahmini Süre:1m 30s

Soru 13Soru →

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

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

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}}

.

Adım Adım Çözüm

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.

Anahtar Kavram

Simplification of nested radical expressions using fractional exponent laws

Soru 14Soru →

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

Cevabı ve açıklamayı göster

Cevap: $9y^2$

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

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

.

Tahmini Süre:45s

Soru 15Soru →

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

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Exponent rules: Power of a Product

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

and Quotient of Powers

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

Soru 16Soru →

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}

Cevabı ve açıklamayı göster

Cevap: $3^5$

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Soru 17Soru →

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

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving Radical Equations and Eliminating Extraneous Solutions

Tahmini Süre:2m 0s

Soru 18Soru →

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

Cevabı ve açıklamayı göster

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

Cevap

\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}

.

Adım Adım Çözüm

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

.

Anahtar Kavram

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

Tahmini Süre:2m 0s

Soru 19Soru →

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 $\frac{2(\sqrt[4]{x}^3 + 1)}{\sqrt{x} - 1}$ .

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

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}

.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Soru 20Soru →

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

Cevabı ve açıklamayı göster

Cevap: 52

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Algebraic identities with fractional exponents

Tahmini Süre:2m 0s

Exponents, Radicals, and Algebraic Expressions

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