Algebraic Exponents and Radicals Alıştırma Soruları — GRE General Test — Sayfa 2

Soru 21Soru →

If $x$ and $y$ are real numbers greater than $1$ such that $x^{\sqrt{y}} = y^{\sqrt{x}}$ and $x^3 = y^2$ , what is the value of $x$ ?

\frac{81}{16}

\frac{9}{4}

\frac{27}{8}

\frac{729}{64}

\frac{243}{32}

Cevabı ve açıklamayı göster

Cevap: $\frac{81}{16}$

Cevap

The correct answer is

\frac{81}{16}

.

Expressing

y

as

x^{3/2}

and substituting it into

x^{\sqrt{y}} = y^{\sqrt{x}}

transforms the equation into

x^{x^{3/4}} = x^{\frac{3}{2}x^{1/2}}

. Equating exponents yields

x^{3/4} = \frac{3}{2}x^{1/2}

, which simplifies to

x^{1/4} = \frac{3}{2}

. Raising both sides to the fourth power gives

x = \frac{81}{16}

.

Adım Adım Çözüm

1

Express

y

in terms of

x

using the second given equation.

Since

x > 1

and

y > 1

, taking the square root of both sides of

y^2 = x^3

gives

y = x^{3/2}

.

Converting

y

to an exponential expression of

x

allows single-variable substitution into the first equation.

2

Substitute

y = x^{3/2}

into the first equation

x^{\sqrt{y}} = y^{\sqrt{x}}

.

Note that

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

. Thus, the left side becomes

x^{x^{3/4}}

, and the right side becomes

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

.

Applying exponent rules

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

simplifies both sides to base

x

expressions.

3

Equate the exponents since the bases are equal and greater than 1.

x^{3/4} = \frac{3}{2}x^{1/2}

.

If

x^a = x^b

for

x > 1

, then

a = b

.

4

Divide both sides by

x^{1/2}

to isolate the power of

x

.

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

.

Using the quotient rule for exponents,

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

where

3/4 - 1/2 = 1/4

.

5

Raise both sides to the 4th power to solve for

x

.

x = \left(\frac{3}{2}\right)^4 = \frac{3^4}{2^4} = \frac{81}{16}

.

Raising

(x^{1/4})^4

eliminates the fractional exponent to give

x

.

Anahtar Kavram

Solving systems of exponential equations using fractional exponent rules and base equality properties.

Soru 22Soru →

If $x$ and $y$ are positive real numbers such that $\left(\frac{x^{-1}y^{\frac{2}{3}}}{\sqrt{x^3 y^{-2}}}\right)^{-\frac{3}{4}} = x^a y^b$ , what is the value of $a + b$ ?

\frac{5}{8}

-\frac{5}{8}

\frac{17}{8}

-\frac{1}{8}

-\frac{13}{8}

Cevabı ve açıklamayı göster

Cevap: $\frac{5}{8}$

Cevap

The value of

a + b

is

\frac{5}{8}

.

Simplifying the fraction inside the parentheses gives

x^{-\frac{5}{2}} y^{\frac{5}{3}}

. Raising this to the power of

-\frac{3}{4}

produces

x^{\frac{15}{8}} y^{-\frac{5}{4}}

. Equating exponents yields

a = \frac{15}{8}

and

b = -\frac{5}{4} = -\frac{10}{8}

. Adding these together gives

a + b = \frac{5}{8}

.

Adım Adım Çözüm

1

Rewrite the radical expression in the denominator using fractional exponents.

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

Applying the power rule

(u^v)^w = u^{v \cdot w}

to radical expressions converts square roots to fractional exponents of

\frac{1}{2}

.

2

Simplify the expression inside the main parentheses by combining like base exponents.

\frac{x^{-1}y^{\frac{2}{3}}}{x^{\frac{3}{2}} y^{-1}} = x^{-1 - \frac{3}{2}} y^{\frac{2}{3} - (-1)} = x^{-\frac{5}{2}} y^{\frac{5}{3}}

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

3

Apply the outer exponent

-\frac{3}{4}

to each factor inside the parentheses.

(x^{-\frac{5}{2}} y^{\frac{5}{3}})^{-\frac{3}{4}} = x^{(-\frac{5}{2})(-\frac{3}{4})} y^{(\frac{5}{3})(-\frac{3}{4})} = x^{\frac{15}{8}} y^{-\frac{5}{4}}

According to exponent rules,

(u \cdot v)^p = u^p \cdot v^p

and

(u^p)^q = u^{p \cdot q}

.

4

Identify

a

and

b

and calculate their sum

a + b

.

a = \frac{15}{8}, \quad b = -\frac{5}{4} = -\frac{10}{8} \implies a + b = \frac{15}{8} - \frac{10}{8} = \frac{5}{8}

Match the simplified expression with

x^a y^b

and add the resulting fractional exponents using a common denominator.

Anahtar Kavram

Simplifying nested algebraic expressions with negative and rational exponents using fundamental laws of exponents.

Tahmini Süre:2m 0s

Soru 23Soru →

What is the sum of all real solutions to the equation $\sqrt{3x + 10} - x = 2$ ?

2

-1

-3

5

3

Cevabı ve açıklamayı göster

Cevap: $2$

Cevap

The sum of all real solutions is

2

.

To solve

\sqrt{3x + 10} - x = 2

, isolate the radical to get

\sqrt{3x + 10} = x + 2

. Squaring both sides yields

3x + 10 = x^2 + 4x + 4

. Rearranging into standard quadratic form gives

x^2 + x - 6 = 0

, which factors as

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

, yielding candidates

x = 2

and

x = -3

. Substituting

x = 2

into the original equation yields

\sqrt{16} - 2 = 2

, which is true. Substituting

x = -3

yields

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

, so

x = -3

is extraneous. The only valid solution is

x = 2

, so the sum of all valid solutions is

2

.

Adım Adım Çözüm

1

Isolate the radical expression on one side of the equation

\sqrt{3x + 10} = x + 2

Isolating the radical allows squaring both sides cleanly to eliminate the radical sign.

2

Square both sides of the equation

3x + 10 = (x + 2)^2 = x^2 + 4x + 4

Squaring eliminates the square root, converting the equation into a polynomial form.

3

Rearrange terms into standard quadratic form

ax^2 + bx + c = 0

x^2 + x - 6 = 0

Grouping all terms on one side sets up the equation for factoring.

4

Factor the quadratic equation to find candidate solutions

(x + 3)(x - 2) = 0 \implies x = -3 \text{ or } x = 2

Factoring determines the values of

x

that solve the algebraic polynomial.

5

Test candidate solutions in the original radical equation to filter extraneous roots

For

x = 2

:

\sqrt{3(2) + 10} - 2 = \sqrt{16} - 2 = 4 - 2 = 2

(Valid). For

x = -3

:

\sqrt{3(-3) + 10} - (-3) = \sqrt{1} + 3 = 4 \neq 2

(Extraneous). Thus,

x = 2

is the only valid solution.

Squaring an equation can introduce extraneous roots that do not satisfy the original principal root definition.

Anahtar Kavram

Solving Radical Equations and Filtering Extraneous Roots

Tahmini Süre:1m 30s

Soru 24Soru →

Let $P$ be the product of all positive real numbers $x$ that satisfy the exponential equation $x^{\sqrt{x}} = \left(x^2\sqrt{x}\right)^x$ . What is the value of $P$ ?

\frac{4}{25}

\frac{2}{5}

\frac{25}{4}

\frac{16}{625}

1

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{25}$

Cevap

\frac{4}{25}

Simplifying the right-hand side yields

x^{\sqrt{x}} = x^{\frac{5}{2}x}

. For

x = 1

, both sides equal 1, making

x = 1

a valid solution. For

x > 0

and

x \neq 1

, equating the exponents gives

\sqrt{x} = \frac{5}{2}x

, which reduces to

\sqrt{x} = \frac{2}{5}

, so

x = \frac{4}{25}

. Multiplying all valid solutions together yields

1 \cdot \frac{4}{25} = \frac{4}{25}

.

Adım Adım Çözüm

1

Simplify the right-hand side of the equation using fractional exponent rules.

Since

x^2\sqrt{x} = x^2 \cdot x^{1/2} = x^{5/2}

, the equation becomes

x^{\sqrt{x}} = \left(x^{5/2}\right)^x = x^{\frac{5}{2}x}

.

Combining terms with the same base into a single exponent simplifies comparison between both sides.

2

Check for the base root

x = 1

.

Substituting

x = 1

gives

1^{\sqrt{1}} = 1^1 = 1

and

(1^2\sqrt{1})^1 = 1^1 = 1

. Thus,

x = 1

is a valid solution.

For any exponential equation of the form

x^{f(x)} = x^{g(x)}

,

x = 1

is always a candidate solution because

1^a = 1^b = 1

for all real exponents.

3

Equate the exponents for positive real solutions where

x \neq 1

.

Setting the exponents equal gives

\sqrt{x} = \frac{5}{2}x

.

When the base

x > 0

and

x \neq 1

,

x^{f(x)} = x^{g(x)}

implies

f(x) = g(x)

.

4

Solve the resulting radical equation for

x

.

Divide both sides by

\sqrt{x}

(since

x > 0

):

1 = \frac{5}{2}\sqrt{x} \implies \sqrt{x} = \frac{2}{5} \implies x = \left(\frac{2}{5}\right)^2 = \frac{4}{25}

.

Isolating

\sqrt{x}

and squaring both sides gives the non-trivial solution.

5

Compute the product

P

of all positive real solutions.

P = 1 \cdot \frac{4}{25} = \frac{4}{25}

.

The question requests the product of all positive real values of

x

satisfying the original equation.

Anahtar Kavram

Solving exponential equations with variable bases and radical powers

Soru 25Soru →

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

\frac{5\sqrt{2}}{3}

\frac{4}{3}

2

\frac{5\sqrt{6}}{6}

\frac{5}{3}

Cevabı ve açıklamayı göster

Cevap: $\frac{5\sqrt{2}}{3}$

Cevap

\frac{5\sqrt{2}}{3}

Squaring the given relationship

a^x - a^{-x} = 2

gives

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

, which simplifies to

a^{2x} + a^{-2x} = 6

. Adding 4 to both sides gives

(a^x + a^{-x})^2 = 8

, so

a^x + a^{-x} = 2\sqrt{2}

. By the sum of cubes identity,

a^{3x} + a^{-3x} = (a^x + a^{-x})(a^{2x} + a^{-2x} - 1) = 2\sqrt{2}(6 - 1) = 10\sqrt{2}

. Dividing

10\sqrt{2}

by

6

gives the simplified result

\frac{5\sqrt{2}}{3}

.

Adım Adım Çözüm

1

Square both sides of the given equation

a^x - a^{-x} = 2

.

(a^x - a^{-x})^2 = a^{2x} - 2(a^x)(a^{-x}) + a^{-2x} = 4 \implies a^{2x} + a^{-2x} = 6

.

Expanding the binomial square allows us to find the denominator

a^{2x} + a^{-2x}

directly.

2

Determine the value of

a^x + a^{-x}

.

(a^x + a^{-x})^2 = a^{2x} + 2 + a^{-2x} = 6 + 2 = 8 \implies a^x + a^{-x} = \sqrt{8} = 2\sqrt{2}

.

Since

a > 1

,

a^x > 0

and

a^{-x} > 0

, their sum must be positive.

3

Use the sum of cubes identity

u^3 + v^3 = (u + v)(u^2 - uv + v^2)

to evaluate

a^{3x} + a^{-3x}

.

a^{3x} + a^{-3x} = (a^x + a^{-x})(a^{2x} - 1 + a^{-2x}) = (2\sqrt{2})(6 - 1) = 10\sqrt{2}

.

Factoring the numerator breaks it into terms whose numerical values are known.

4

Compute the ratio of numerator to denominator.

\frac{a^{3x} + a^{-3x}}{a^{2x} + a^{-2x}} = \frac{10\sqrt{2}}{6} = \frac{5\sqrt{2}}{3}.

Simplify the fraction by dividing numerator and denominator by 2.

Anahtar Kavram

Algebraic transformations of exponential expressions using polynomial identities.

Tahmini Süre:2m 30s

Soru 26Soru →

If

x

is a real number greater than

1

satisfying the exponential equation

\left(x^x\right)^{\sqrt{x}} = x^{\left(x^{\sqrt{x}}\right)}

what is the value of

x

?

\frac{3}{2}

\frac{4}{9}

\frac{9}{4}

\frac{27}{8}

\frac{81}{16}

Cevabı ve açıklamayı göster

Cevap: $\frac{9}{4}$

Cevap

The value of

x

is

\frac{9}{4}

.

Applying the exponent rule

(a^b)^c = a^{bc}

simplifies the left side to

x^{x\sqrt{x}}

. Since the base

x > 1

is identical on both sides, we set the exponents equal:

x\sqrt{x} = x^{\sqrt{x}}

. Expressing

x\sqrt{x}

as

x^{3/2}

yields

x^{3/2} = x^{\sqrt{x}}

, which implies

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

. Squaring both sides gives

x = \frac{9}{4}

.

Adım Adım Çözüm

1

Apply the power rule

(a^b)^c = a^{bc}

to the left-hand side of the equation.

\left(x^x\right)^{\sqrt{x}} = x^{x \cdot \sqrt{x}} = x^{x^{3/2}}

Raising a power to another exponent requires multiplying the exponents:

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

.

2

Equate the exponents of the expressions on both sides, as the bases are equal and

x > 1

.

x \sqrt{x} = x^{\sqrt{x}}

, which means

x^{3/2} = x^{\sqrt{x}}

If

x^A = x^B

and

x > 1

, then

A = B

.

3

Equate exponents once more for the base

x

.

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

Since the bases are identical (

x > 1

), their exponents must be equal.

4

Square both sides to solve for

x

.

x = \left(\frac{3}{2}\right)^2 = \frac{9}{4}

Squaring

\sqrt{x}

isolates

x

.

Anahtar Kavram

Properties of exponents and nested power rules with radicals

Tahmini Süre:2m 0s

Soru 27Soru →

If

x

is a positive real number satisfying the equation

\sqrt[3]{\frac{x^3 \sqrt{x\sqrt{x}}}{x^{-1/4}}} = 16

what is the value of

x

?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8

By converting all radicals into fractional exponents and systematically applying exponent rules, the expression under the cube root simplifies to

x^4

. Taking the cube root gives

x^{4/3} = 16

. Solving for

x

by raising both sides to

3/4

yields

x = 16^{3/4} = (2^4)^{3/4} = 2^3 = 8

.

Adım Adım Çözüm

1

Express the inner nested radical using fractional exponents

\sqrt{x\sqrt{x}} = \sqrt{x \cdot x^{1/2}} = \sqrt{x^{3/2}} = x^{3/4}

Applying the product and power rules of exponents:

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

and

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

.

2

Simplify the numerator inside the outer radical

x^3 \cdot x^{3/4} = x^{3 + 3/4} = x^{15/4}

Multiplying exponential terms with the same base requires adding their exponents.

3

Divide by the negative exponent in the denominator

\frac{x^{15/4}}{x^{-1/4}} = x^{15/4 - (-1/4)} = x^{16/4} = x^4

Dividing exponential terms with the same base requires subtracting the denominator exponent from the numerator exponent.

4

Apply the outer cube root to the simplified expression

\sqrt[3]{x^4} = (x^4)^{1/3} = x^{4/3}

The

n

-th root of an expression is equivalent to raising that expression to the power of

1/n

.

5

Solve the resulting exponential equation for

x

x^{4/3} = 16 \implies x = 16^{3/4} = (2^4)^{3/4} = 2^3 = 8

Raise both sides of

x^{4/3} = 16

to the power of

3/4

to isolate

x

.

Anahtar Kavram

Simplifying nested algebraic radicals and solving equations with fractional exponents using exponent rules.

Soru 28Soru →

If $x$ is a real number satisfying the radical equation $\sqrt{x + 7} - x = 1$ , what is the value of $x$ ?

-3

-2

2

3

6

Cevabı ve açıklamayı göster

Cevap: $2$

Cevap

The value of

x

is

2

.

Isolating the square root yields

\sqrt{x + 7} = x + 1

. Squaring both sides produces

x + 7 = x^2 + 2x + 1

, which reduces to the quadratic equation

x^2 + x - 6 = 0

. Factoring yields

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

, giving potential roots of

x = -3

and

x = 2

. Testing

x = 2

in the original expression yields

\sqrt{2+7} - 2 = 3 - 2 = 1

, confirming it is correct.

Adım Adım Çözüm

1

Isolate the radical expression on one side of the equation.

\sqrt{x + 7} = x + 1

Isolating the radical allows squaring both sides cleanly to eliminate the radical sign.

2

Square both sides of the equation.

x + 7 = (x + 1)^2 = x^2 + 2x + 1

Squaring removes the square root on the left side.

3

Rearrange into standard quadratic form

ax^2 + bx + c = 0

.

x^2 + x - 6 = 0

Subtracting

x + 7

from both sides collects all terms on one side.

4

Factor the quadratic equation and solve for potential values of

x

.

(x + 3)(x - 2) = 0 \implies x = -3 \text{ or } x = 2

Setting each factor to zero identifies potential solutions.

5

Substitute each potential solution back into the original equation

\sqrt{x + 7} - x = 1

to check for extraneous roots.

For

x = 2

:

\sqrt{2 + 7} - 2 = 3 - 2 = 1

(Valid). For

x = -3

:

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

(Extraneous).

Squaring an equation can introduce false solutions that must be eliminated.

Anahtar Kavram

Solving radical equations and checking for extraneous solutions

Tahmini Süre:1m 30s

Soru 29Soru →

For all positive real numbers $a$ and $b$ , which of the following expressions are equivalent to $\left( \frac{a^{-2} b^3}{\sqrt{a^4 b^{-2}}} \right)^{-1/2}$ ? Select all that apply.

Geçerli olan tümünü seçin

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

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $\frac{a^2}{b^2}$ ; $\left( \frac{b}{a} \right)^{-2}$ ; $\sqrt{\frac{a^4}{b^4}}$

Cevap

The expressions equivalent to the given quantity are

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

,

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

, and

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

.

Simplifying the original expression step-by-step yields

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

. The expressions

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

,

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

, and

\sqrt{\frac{a^4}{b^4}} = \frac{a^2}{b^2}

are all identical to the simplified form.

Adım Adım Çözüm

1

Simplify the radical in the denominator of the inner expression.

Since

a > 0

and

b > 0

,

\sqrt{a^4 b^{-2}} = \sqrt{a^4} \cdot \sqrt{b^{-2}} = a^2 b^{-1}

.

Applying the square root to each variable power individually.

2

Simplify the expression inside the outer parenthesis.

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

Subtracting exponents of like bases according to the quotient rule of exponents.

3

Apply the outer exponent of

-\frac{1}{2}

.

(a^{-4} b^4)^{-1/2} = (a^{-4})^{-1/2} (b^4)^{-1/2} = a^2 b^{-2} = \frac{a^2}{b^2}.

Multiplying internal exponents by

-\frac{1}{2}

according to the power rule of exponents.

4

Evaluate each choice against the simplified form

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

.

The expressions

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

,

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

, and

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

are all algebraically equivalent to

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

.

Testing algebraic equivalence using standard exponent and radical laws.

Anahtar Kavram

Properties of exponents and radicals, including power of a quotient, quotient rule, negative exponents, and square roots of powers.

Tahmini Süre:1m 30s

Soru 30Soru →

For all real numbers $x$ , what is the numerical value of the expression $\frac{3^{x+2} - 3^x}{3^{x-1} + 3^{x-2}}$ ?

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

The numerical value of the expression is 18.

Factoring

3^x

from both terms in the numerator gives

3^x(3^2 - 1) = 8 \cdot 3^x

. Factoring

3^x

from both terms in the denominator gives

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

. Dividing the two expressions cancels

3^x

entirely, resulting in

\frac{8}{\frac{4}{9}} = 8 \times \frac{9}{4} = 18

.

Adım Adım Çözüm

1

Factor out

3^x

from the numerator.

3^{x+2} - 3^x = 3^x(3^2 - 1) = 3^x(9 - 1) = 8 \cdot 3^x

Applying the exponent rule

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

allows factoring out the common exponential factor

3^x

.

2

Factor out

3^x

from the denominator.

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

Applying negative exponent rules

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

allows evaluating the remaining numerical terms inside the parentheses.

3

Simplify the overall fraction by dividing the factored numerator by the factored denominator.

\frac{8 \cdot 3^x}{\frac{4}{9} \cdot 3^x} = \frac{8}{\frac{4}{9}} = 8 \times \frac{9}{4} = 18

The non-zero common term

3^x

cancels from both numerator and denominator, leaving a constant integer.

Anahtar Kavram

Factoring and simplifying exponential expressions with variable exponents.

Soru 31Soru →

If $x$ is a real number satisfying the exponential equation $9^{x+1} - 3^{2x+1} = 162$ , what is the value of $4^x$ ?

4

6

8

16

64

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8

By converting

9^{x+1}

to

3^{2x+2}

and factoring out

3^{2x}

, the equation simplifies to

6 \cdot 3^{2x} = 162

. Dividing by 6 gives

3^{2x} = 27

, so

2x = 3

and

x = \frac{3}{2}

. Raising 4 to the power of

\frac{3}{2}

yields

(\sqrt{4})^3 = 8

.

Adım Adım Çözüm

1

Rewrite terms with a common base of 3.

9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2}

Since 9 is a power of 3 (

9 = 3^2

), applying exponent rules converts the equation to base 3.

2

Factor out the common exponential expression

3^{2x}

.

3^{2x+2} - 3^{2x+1} = 3^{2x} \cdot 3^2 - 3^{2x} \cdot 3^1 = 3^{2x}(9 - 3) = 6 \cdot 3^{2x}

Using product rule of exponents (

3^{a+b} = 3^a \cdot 3^b

) allows combining like terms.

3

Solve for

x

.

6 \cdot 3^{2x} = 162 \implies 3^{2x} = 27 \implies 3^{2x} = 3^3 \implies 2x = 3 \implies x = \frac{3}{2}

Dividing both sides by 6 yields

3^{2x} = 27

, and equating exponents of matching bases gives

x = \frac{3}{2}

.

4

Evaluate the target expression

4^x

.

4^{3/2} = (4^{1/2})^3 = 2^3 = 8

Substituting

x = \frac{3}{2}

into

4^x

means taking the square root of 4 and raising it to the third power.

Anahtar Kavram

Solving exponential equations using common bases and exponent properties

Algebraic Exponents and Radicals

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