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Question : 31 of 86

Marks: +1, -0

The value of

x

x^{x\sqrt{x}} = (x\sqrt{x})^{x}

x

$\frac{4}{9}$
$\frac{9}{4}$
$-\frac{9}{4}$
none

Solution:

Concept:

Simplify exponents using laws of indices and equate powers.

Explanation:

Note that

x\sqrt{x} = x \cdot x^{1/2} = x^{3/2}

.

The equation becomes

x^{x^{3/2}} = (x^{3/2})^x = x^{(3/2)x}

.

For

x=1

, both sides equal

1

, so

x=1

is a solution.

For

x>0

and

x\neq 1

, equate exponents:

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

.

Divide by

x

(positive):

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

.

Square:

x = \left(\frac{3}{2}\right)^2 = \frac{9}{4}

.

Answer:

x = \frac{9}{4}

(Option B)

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