# 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)

• I think you are misreading the question. I'm pretty sure that most of those expressions on the RHS are supposed to be indices of the radicals, i.e., $\sqrt[x^2]{\sqrt[x(x-1)]{\cdots }}$, so if $x=5$ then $\sqrt{\sqrt{\cdots }}$ – PM 2Ring Jun 19 '16 at 9:28
• @PM2Ring Really ?? Hmmm I didn't see that way, okay let me try that way – Arnav Das Jun 19 '16 at 9:39
• You have a typo in RHS (first line): It is not $2x\sqrt{x\sqrt x}$ but $\sqrt[2x]{\sqrt[x]{\sqrt x}}$ – Piquito Jun 19 '16 at 14:20
• @Piquito could you solve it then, i am at a loss to solve that way too – Arnav Das Jun 21 '16 at 13:20