If x and y are positive real numbers satisfying the system of exponential radical equation… — GRE General Test

x

and

y

are positive real numbers satisfying the system of exponential radical equations

\sqrt{x\sqrt{y}} = 3

\sqrt{y\sqrt{x}} = 9

which of the following statements must be true? Select all that apply.

$xy = 81$ Answer
$x^{1/2} + y^{1/2} = 10$ Answer
C
$\sqrt{xy} = 27$
D
$x$ has two valid real solutions, $x = 1$ and $x = -1$
$y^x - x^y = 80$ Answer

Answer

The statements asserting that

xy = 81

x^{1/2} + y^{1/2} = 10

, and

y^x - x^y = 80

are true.

Solving the system of radical equations gives the unique positive solution pair

x = 1

and

y = 81

. Substituting these values into the statements shows that

xy = 1 \cdot 81 = 81

x^{1/2} + y^{1/2} = \sqrt{1} + \sqrt{81} = 10

, and

y^x - x^y = 81^1 - 1^{81} = 80

are all mathematically valid.

Step-by-Step Solution

Square both sides of each equation to remove the outer radicals

x\sqrt{y} = 9 \implies x^2 y = 81

and

y\sqrt{x} = 81 \implies y^2 x = 6561

Squaring

\sqrt{A} = B

yields

A = B^2

, and rewriting radicals as rational powers gives

x y^{1/2} = 9

and

y x^{1/2} = 81

Express

y

in terms of

x

from the first equation and substitute into the second

y = \frac{81}{x^2} \implies \left(\frac{81}{x^2}\right)^2 x = 6561 \implies \frac{6561}{x^3} = 6561

Substituting

y

eliminates the variable

y

, leaving a single equation in terms of

x

Solve for

x

and

y

x^3 = 1 \implies x = 1

, which gives

y = \frac{81}{1^2} = 81

Since

x > 0

is a positive real number,

x = 1

is the unique real solution, yielding

y = 81

Evaluate each given statement using

x = 1

and

y = 81

xy = 81

(True),

x^{1/2} + y^{1/2} = 1 + 9 = 10

(True),

\sqrt{xy} = 9 \neq 27

(False),

x = -1

is invalid (False),

81^1 - 1^{81} = 80

(True)

Direct calculation confirms which individual statements hold.

Key Concept

Simplifying nested radical equations using fractional exponent rules

Estimated Time:2m 0s

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Step-by-Step Solution

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