If x and y are real numbers greater than 1 such that x^{\sqrt{y}} = y^{\sqrt{x}} and x^3 =… — GRE General Test

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 $x$ ?

The correct answer is

\frac{81}{16}

Expressing

y

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}

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.

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.

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

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

x^a = x^b

for

x > 1

, then

a = b

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

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

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

Question