# 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 $\sqrt[25]{\sqrt[20]{\cdots }}$ Jun 19, 2016 at 9:28
• @PM2Ring Really ?? Hmmm I didn't see that way, okay let me try that way Jun 19, 2016 at 9:39
• You have a typo in RHS (first line): It is not $2x\sqrt{x\sqrt x}$ but $\sqrt[2x]{\sqrt[x]{\sqrt x}}$ Jun 19, 2016 at 14:20
• @Piquito could you solve it then, i am at a loss to solve that way too Jun 21, 2016 at 13:20

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 $$(x-1)!x^{x+1}=\frac{1}{2}$$.