Cfrgtkky

Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.

There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:

The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.

Then:
begin{eqnarray}
frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
end{eqnarray}
Then with Lambert function:
begin{eqnarray}
-frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
end{eqnarray}
Then the solutions are:
begin{eqnarray}
x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.44836908960ldots \
x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.65205358ldots-i101297.96245405ldots \
x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.45994697086ldots \
end{eqnarray}

The function is numeric too, but it is very elegant.

There is an article in wikipedia for this Lambert W function.

This equation has three real solutions:
$$xapprox 0.4483690898$$
$$x=4$$
$$xapprox 6380.459941$$
This can be obtained by the Newton Raphson method.

Welcome to the world of Lambert function !

Using Matti P.'s suggestion $x=2^y$, the equation becomes
$$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
$$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
$$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
$$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$

It have been faster to use $x=2^{e^t}$ for the same result.