<s>[INST] Does this equation have no solutions? The question is this :

The source from where I got this question was devoid of any answers to it, so I came here, this is how I proceeded :
LHS :
$((((({(x)^x})^{2x})^{3x})^{....x^2})^2 = (((((x)^{2x^2})^{3x})^{....x^2})^2 =...........= (x^{(x^x)x!})^2 = x^{2(x^x)x!} $
RHS :
$\sqrt{x^2\sqrt{(x-1)x\sqrt{(x-2)x\sqrt{...........2x\sqrt{x\sqrt x}}}}} =  \sqrt{x\sqrt{(x-1)\sqrt{(x-2)\sqrt{...........2\sqrt{1\sqrt 1}}}}}*x^{(1-\frac{1}{2^{x+1}})} $ $= x^{1/2}*(x-1)^{1/4}*(x-2)^{1/8}*....... *{2}^{(\frac{1}{2^{x-1}})}* x^{(1-\frac{1}{2^{x+1}})} $
Now I thought of taking $\log$ of both RHS and LHS from which I could deduce
LHS:
$\log x^{2(x^x)x!} = 2(x^x)x!\log x = 2*x^{x+1}(x-1)!\log x $
RHS:
$\log (x^{1/2}*(x-1)^{1/4}*(x-2)^{1/8}*....... *{2}^{(\frac{1}{2^{x-1}})}* x^{(1-\frac{1}{2^{x+1}})})$ $= \frac{\log x}{2}+\frac{\log (x-1)}{4}+ \frac{\log (x-2)}{8}+..........+ \frac{\log 2}{2^{x-1}}+ {(1-\frac{1}{2^{x+1}})} \log x $
Now equating LHS = RHS I get :
$ 2*x^{x+1}(x-1)!\log x = \frac{\log x}{2}+\frac{\log (x-1)}{4}+ \frac{\log (x-2)}{8}+..........+ \frac{\log 2}{2^{x-1}}+ {(1-\frac{1}{2^{x+1}})} \log x$ $\implies x^{x+1}(x-1)! =\frac{ \frac{\log x}{2}+\frac{\log (x-1)}{4} + \frac{\log (x-2)}{8}+..........+\frac{\log 2}{2^{x-1}}+ {(1-\frac{1}{2^{x+1}})} \log x}{2\log x} $
now in the RHS of above equation I only found $\log x$ in 3 places : denominator, first place of numerator and in the last place of numerator; I assumed terms from $\frac {\log (x-1)}{4*2\log x}$ to $\frac{\log 2}{2^{x-1}*2\log x}$ were becoming too small to take into calculation, so final equation could be written down to :
$ x^{x+1}(x-1)! = \frac{ \frac{\log x}{2} + {(1-\frac{1}{2^{x+1}})} \log x}{2\log x} =\frac{\frac{1}{2} +1- \frac{1}{ 2^{x+1} }}{2} = \frac{3}{4} -  \frac{1}{ 2^{x+2} } $$\implies x^{x+1}(x-1)! = \frac{3}{4} -  \frac{1}{ 2^{x+2} }$
the above equation is where I am forced to stop, please guide me after that ? Or did I take a wrong approach from start itself ?
(If you can, do tag it with appropriate tags; I could not find the suitable ones for this problem)
 [/INST] The equation can be written as:
$x^{2x^{x} x!}=\sqrt{x^{\frac{ 1}{x! x^{x}} }}$.
By squared up the two members, we get:
$x^{4x^{x} x!}=x^{\frac{ 1}{x! x^{x}}}$.
Since this is the equality between two powers, for them to be equal they must have the same basis and the same exponent; therefore, since the basics are the same, we can write:
$ 4x^{x} x!= \frac{ 1}{x! x^{x}}$.
Solving for $x^{x} x!$, you get two values $±\frac{1}{2}$.
Noting that the expression $(x-1)!x^{x+1}$, is none other than $x^{x} x!$, as
$(x-1)!x^{x+1}=(x-1)! x x^{x}=x!x^{x}$,
the solution is the positive real value $(x-1)!x^{x+1}=\frac{1}{2}$.
</s>