If x is a positive real number satisfying the equation \sqrt[3]{\frac{x^3 \sqrt{x\sqrt{x}}… — GRE General Test

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

Answer: 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

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}

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.

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.

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

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

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

Question