The number of solutions of the equation (x)^{ | vedclass

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(C) Given equation: $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$Case $1$: If $x = 1$,then $1^{1\sqrt{1}} = 1^1$,which is $1 = 1$. So,$x = 1$ is a solution.Case $

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$2$

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(C) Given equation: $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$Case $1$: If $x = 1$,then $1^{1\sqrt{1}} = 1^1$,which is $1 = 1$. So,$x = 1$ is a solution.Case $2$: If $x > 0$ and $x 
eq 1$,we can write the equation as:$x^{x^{3/2}} = (x^{3/2})^x$$x^{x^{3/2}} = x^{(3/2)x}$Equating the exponents:$x^{3/2} = \frac{3}{2}x$Since $x 
eq 0$,divide by $x$:$x^{1/2} = \frac{3}{2}$$x = (\frac{3}{2})^2 = \frac{9}{4}$Thus,the solutions are $x = 1$ and $x = \frac{9}{4}$.The total number of solutions is $2$.

The number of solutions of the equation $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$ is:

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