how to calculate $a^x=x$ for $a>0$? I'm struggling with this equation, but i think i'm pretty close to the solution, but i just can't figure it out, so i hope for some help!
$$a^x=x \rightarrow a^x-x=0$$
Now i take derivative for $x$ and it is 
$$a^x*\log(a) - 1 = 0$$ 
Checking where derivative is positive.
$$a^x*\log(a) > 1$$ 
And here i'm basically stuck, because i can say, by guessing, that for $a=1,e,e^2,e^3\dots$ $a^x-x$ is increasing, then where it's decreasing, find extremas, and find solutions, but i don't feel like i will get correct answer for this "by guessing"... I would appreciate some help on this! Thanks in advance
 A: Assuming that $a \neq 1$ (if $a=1$ then $x=1$ of course), this equation is equivalent to
$$\begin{align}
x a^{-x} = 1 &\Leftrightarrow x e^{-x \ln(a)} = 1 \\
& \Leftrightarrow (-x \ln(a)) e^{-x \ln(a)} = -\ln(a) \\
& \Leftrightarrow - x \ln(a) = W(-\ln(a)) \\
& \Leftrightarrow x = \color{red}{- \frac{W(-\ln(a))}{\ln(a)}}
\end{align}$$
Where $W$ is the Lambert W function. I don't think there's a "better" solution than that -- W doesn't have a closed form in terms of elementary functions.
A: As answered by nik, there is no analytical solution beside the Lambert function. If you cannot use it, then only numerical methods can be used and Newton is the simplest, provided a reasonable guess for starting the iterative process. For this case, using $$x(n+1)=x(n)-\frac{a^{x(n)}-x(n)}{ a^{x(n)}\log (a)-1}$$  should converge quite fast.  
Just as an example, select $a=\frac{1}{2}$; by inspection, we know that there is a solution between $0$ and $1$. Let me be very lazy and I shall start iterating from $x=0$. The successive iterates will be $0.590616$, $0.640910$ and $0.641186$ which is the solution for six significant figures. Using Lambert function, the solution is $0.641185744504986$.
