Trying to solve $x^2=\frac 1{\ln x}$ Can anyone help me solve $x^2=\dfrac 1{\ln x}$ ? I don't have any idea.
Is there any way to solve it with exponentials? 
I already proved the existence of the solution by IVT. Indeed the function $f(x)=1/x\ln x$ is strictly decreasing in $]1;+\infty[$ with $\lim_{x\to 1^+} f(x)= +\infty$ and $\lim_{x\to \infty} f(x)= 0$, so it has to cross the $y=x$ line once. 
 A: As already said, solve it numerically for 
$$x\simeq 1.531584$$
The analytical solution cannot be expressed with a finite number of elementary functions. It requires a special function, the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html
$$x^2=\frac{1}{\ln(x)}=\frac{2}{\ln(x^2)}$$
$$\ln(x^2)=\frac{2}{x^2}$$
$$x^2=e^{2/x^2}$$
$$\frac{2}{x^2}e^{2/x^2}=2$$
Let $X=\frac{2}{x^2}$
$$Xe^X=2$$
From the definition of the Lambert W function :
$$X=W(2)$$
Thus $\frac{2}{x^2}=W(2)$
$$x=\sqrt{\frac{2}{W(2)}}$$
$W(2)\simeq 0.8526055$
A: The first thing, as you have made, is to show/check that there is exactly a solution for $x>1$ by IVT.
To determine that value by numerical methods (Newton's, bisection, etc.) we can use bisection method by a calculator starting for example from


*

*$x_a=1 \implies f(1)<0$

*$x_b=2 \implies f(2)>0$
and then we can iteretively get closer and closer to the solution by


*

*$x_i=\frac{x_a+x_b}2$

*if $f(x_i)<0 \implies x_a=f(x_i)$ 

*if $f(x_i)>0 \implies x_b=f(x_i)$
Here is the numerical solution by WolframAlpha that is $x \approx 1.5316$.
A: If you do not want to use Lambert function, consider that you look for the zero of function
$$f(x)=\log(x)-\frac 1{x^2}$$ for which
$$f'(x)=\frac{2}{x^3}+\frac{1}{x}> 0 \,\,\,\, \forall x\qquad \text{and} \qquad f''(x)=-\frac{6}{x^4}-\frac{1}{x^2} < 0 \,\,\,\, \forall x$$
The first derivative does not show real roots. By inspection $f(1)=-1$ and $f(2)=\log (2)-\frac{1}{4}$
Build a Taylor series at $x=1$ to get
$$f(x)=-1+3 (x-1)+O\left((x-1)^2\right)$$ giving the approximate solution  $\frac 4 3$. 
Using instead the simplest Padé approximant, you would have
$$f(x)\sim \frac{17-11 x}{1-7 x}$$ giving the approximate solution  $\frac {17}{11}\approx 1.54545 $ which is not too bad compared to the exact value.
Edit
You could have better approximations if you notice that $f\left(\sqrt{e}\right)=\frac{1}{2}-\frac{1}{e}$. So, using Taylor again
$$f(x)=\left(\frac{1}{2}-\frac{1}{e}\right)+\frac{(2+e)
   \left(x-\sqrt{e}\right)}{e^{3/2}}+O\left(\left(x-\sqrt{e}\right)^2\right)$$ which gives as an estimate
$$x=\frac{\sqrt{e} (6+e)}{2 (2+e)}\approx 1.52323$$
Doing the same with the simplest Padé approximant, we should get
$$f(x)\sim\frac{(4+5 e (4+e)) x-\sqrt{e} (6+e) (2+3 e)}{2 e (6+e) x+2 (e-2) e^{3/2}}$$ giving as an estimate
$$x=\frac{\sqrt{e} (6+e) (2+3 e)}{4+5 e (4+e)}\approx 1.53147$$
