Intersection of two x powers Many months ago in class I came up with the problem: $$x^{(x+1)} = (x+1)^x$$
Using the solve function on my calculator I have found that the answer is around 2.29... This is backed up by the graph. However I was determined to find the inverse function where: $$x = f(y) $$ or find the answer algebraically $$ x = _-$$ Being a lowly first year A-level student this has been pretty much impossible. So far doing some simple rearranging the equation looks like: $$ x^{\frac 1 x} - \frac 1 x - 1=0$$I've tried many methods and had a good look online. So far I have just about been able to solve the equation $x^{\frac 1 x} = y$ by finding the inverse of $x^x$ graphically so $$ x^x = y $$ $$ x = P(y)$$
 where P is the inverse function of $x^x$, then doing $$x^{\frac 1 x} = y$$ $$e^{ln(x^{\frac 1 x})}=y$$ $$\frac1 x (ln(x))=ln(y)$$ $$xln(x)=\frac 1 {ln(y)}$$ $$x^x = e^{\frac1 {ln(y)}}$$ $$x=P\biggl(e^{\frac1 {ln(y)}}\biggr)$$I don't know how to fit this in to my original equation to have just x on one side and no x's on the other side ... I do not want to use any guesswork methods or methods where you work your way to the answer slowly. I have tried using methods where you go one step up above powers so $x^x$ becomes something like $x@2$ where @ is used like + or X then trying to find the inverse of this like - is to + and / is to X  and $\sqrt x$ is to $x^2$, to help you bridge the barrier between the $x^x$ and the $\frac 1 x$ but I couldn't find any way of doing this. Thank you for the help. 

We can approximate the solution building around $x=2$ the Taylor expansion of function $$f(x)=(x+1)\log(x)-x \log(x+1)$$ the first derivatives of which being$$f'(x)=\frac{1}{x}+\frac{1}{x+1}+\log (x)-\log (x+1)$$
 $$f''(x)=-\frac{x^2+x+1}{x^2 (x+1)^2}\qquad f'''(x)=\frac{(2 x+1) \left(x^2+x+2\right)}{x^3 (x+1)^3}$$ This would give
$$f(x)=-\log
   \left(\frac{9}{8}\right)+ \left(\frac{5}{6}-\log \left(\frac{3}{2}\right)\right)(x-2)-\frac{7}{72}
   (x-2)^2+\frac{5}{162} (x-2)^3+O\left((x-2)^4\right)$$
 Using the expansion to $O\left((x-2)^2\right)$ would give a solution which is $$x=2+\frac{6 \log \left(\frac{9}{8}\right)}{5-\log \left(\frac{729}{64}\right)}\approx \color{red}{2.2}7528$$ I shall not write the solution of the quadratic corresponding to the expansion up to $O\left((x-2)^3\right)$ but the solution will be $x\approx \color{red}{2.29}506$.
Solving the cubic corresponding to the expansion up to $O\left((x-2)^4\right)$ would give $x\approx \color{red}{2.29}297$.
We could even do better building the simplest $[1,1]$ Padé approximant of function $f(x)$ 
$$f(x) \approx \frac{f(2)+\left(f'(2)-\frac{f(2) f''(2)}{2 f'(2)}\right)(x-2)  } {1-\frac{ f''(2)}{2 f'(2)}(x-2) }$$ and solving for $x$ the numerator get an explicit expression which evalutes $x\approx \color{red}{2.293}65$
Building the simplest $[1,2]$ Padé approximant would lead to  $x\approx \color{red}{2.2931}3$ still a the price of a linear equation.
