Let P be the product of all positive real numbers x that satisfy the exponential equation … — GRE General Test

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 $P$ ?

$\frac{4}{25}$ Answer
B
$\frac{2}{5}$
C
$\frac{25}{4}$
D
$\frac{16}{625}$
E
$1$

Answer

\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}

Step-by-Step Solution

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.

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.

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)

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.

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.

Key Concept

Solving exponential equations with variable bases and radical powers

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