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$x\sqrt{x}logx = \frac{3}{2}xlogx$

$xlogx(\sqrt{x} - \frac{3}{2}) = 0$

$x$ can't be zero , thus 2 solutions

Expand the right-hand side to get

$x^{x\sqrt{x}}=x^{x}\times \sqrt{x}^{x}$

We are told that x is not 0,

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

First take log on both sides...then on squaring we get x^3=9x^2/4... We can easily do the question by plotting curves between y=x^3 and y=9x^2/4 hence we will get 3 points of intersection out of which zero is not in the domain . hence only two possible solutions

$x^{x*\sqrt{x}} =x^{x^{3/2}}$
$(x*\sqrt{x})^{x}=(x^x*x^{x/2} ) = x^{(3/2)x}$
$\therefore ~x=1~~~OR~~~ idex~~ to ~the ~same~ base..... x^{3/2} = (3/2)x \\ ~~~~~~~~~\therefore ~~x(\sqrt{x} - 3/2)=0......if x \neq0.$
x can not be equal to 0. 0 can not be raised to any power.
So two numbers are 1 and 9/4.
(Please note, I have made correction.)

1 for log x 9/4 for( x^1/2)-3/2

Take log of both sides to reduce the equation to :

x * Sqrt(x) log(x)=(3/2) log(x)

Thus, x * Sqrt(x)=(3/2), which implies Square root of x= 3/2 .

This gives x=9/4 which is one possible solution and the other one is 1.

2 solutions .. because you can substitute the x by 1 or 0