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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 $f(x)=1/xln x$ is strictly decreasing in $]1;+infty[$ with $lim_{xto 1^+} f(x)= +infty$ and $lim_{xto infty} f(x)= 0$, so it has to cross the $y=x$ line once.

As already said, solve it numerically for
$$xsimeq 1.531584$$
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 $frac{2}{x^2}=W(2)$

$$x=sqrt{frac{2}{W(2)}}$$
$W(2)simeq 0.8526055$

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 $x approx 1.5316$.

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
$$f'(x)=frac{2}{x^3}+frac{1}{x}> 0 ,,,, forall xqquad text{and} qquad f''(x)=-frac{6}{x^4}-frac{1}{x^2} < 0 ,,,, forall x$$
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
$$f(x)=-1+3 (x-1)+Oleft((x-1)^2right)$$ 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 $fleft(sqrt{e}right)=frac{1}{2}-frac{1}{e}$. So, using Taylor again
$$f(x)=left(frac{1}{2}-frac{1}{e}right)+frac{(2+e)
left(x-sqrt{e}right)}{e^{3/2}}+Oleft(left(x-sqrt{e}right)^2right)$$ 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
$$f(x)simfrac{(4+5 e (4+e)) x-sqrt{e} (6+e) (2+3 e)}{2 e (6+e) x+2 (e-2) e^{3/2}}$$ giving as an estimate
$$x=frac{sqrt{e} (6+e) (2+3 e)}{4+5 e (4+e)}approx 1.53147$$