If x > 1 and \sqrt{x^{\sqrt{x}}} = x^2, what is the value of x? — GRE General Test

If $x > 1$ and $\sqrt{x^{\sqrt{x}}} = x^2$ , what is the value of $x$ ?

By rewriting the square root as an exponent of

1/2

, the left side becomes

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

. Since the base

x

is greater than 1, set the exponents equal to each other:

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

, which gives

\sqrt{x} = 4

. Squaring both sides yields

x = 16

Express the radical on the left side of the equation as a fractional exponent.

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

The square root rule states that

\sqrt[k]{a} = a^{\frac{1}{k}}

Set the exponent of the left side equal to the exponent of the right side.

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

Since the bases are equal (

x > 1

), their corresponding exponents must be equal.

Solve for

\sqrt{x}

by multiplying both sides by 2.

\sqrt{x} = 4

Isolating the radical term allows determination of the root value.

Square both sides to solve for

x

x = 4^2 = 16

Squaring a principal square root yields the underlying radicand.

Combining Fractional Exponents and Radical Expressions

Question