Question

pls friends give me the solution of this question. Tommorow my exam is there

Answer 1

Given Equation is:

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

Apply log on both sides, we get

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

$\sqrt{x} log x = x log( \sqrt{x} )$   ------------------------- (1)

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

$\sqrt{x} log x = x * \frac{1}{2} log x$

$\sqrt{x} log x - x * \frac{1}{2} log x = 0$

$logx(- \frac{1}{2} x + \sqrt{x} ) = 0$

Here we are having 2 equations,

(1)

 log x = 0

 log x = log 1

 x = 1


(2)

$\frac{-1}{2} x + \sqrt{x} = 0$

$-x + 2 \sqrt{x} = 0$

$2 \sqrt{x} = x$

On squaring both sides, we get

4x = x^2

x^2 - 4x = 0

x(x - 4) = 0

x = 0,4.


Therefore the equations are x = 1,0,4.


Substitute x = 1,x = 4 and x = 0 in (1), we get

$\sqrt{1} log(1) = 1log( \sqrt{1} )$  ---- This is a true statement.

$\sqrt{4} log 4 = 4 log \sqrt{4}$   ---- Ture statement.

$\sqrt{0} log 0 = 0 log( \sqrt{0} )$  ------- False statement.


Therefore the required solutions are 1 and 4.


Hope this helps!