Fun logarithm question How would you go about solving $x^{2^{x}} = 2^{x}$? There should be a solution $1&lt;x&lt;2$, but I haven't found a way to derive the answer using the usual log laws, maybe there is an elegant way though...
We all agree on the fact that the solution of this equation is close to $e^{1/e}$. What I did was to write the solution as $x=e^{1/(e+y)}$ and I expanded the equation as a Taylor series around $y=0$. Limited to the first order, this leads to an expression of $y$ (I am physically unable to write it down) which only includes $2$, e and $\log 2$. Its numerical value is $2.53791*10^{-6}$. For this value of $x=1.444667364812765575108917$ (quite close to Umberto's), the function value is $5.48006*10^{-12}$.
Edit (6 years later)
Taking logarithms, consider that we look for the zero of function
$$f(x)=\log \left(x^{2^x}\right)-\log \left(2^x\right)=2^x\log(x)-x \log(2)$$ and perform one Newton iteration to get
$$x_0=e^{\frac{1}{e}}\implies x_1=x_0+   
   \frac{{x_0} \left({x_0} \log (2)-2^{x_0} \log ({x_0})\right)}{2^{x_0}+2^{x_0} {x_0} \log (2) \log ({x_0})-{x_0} \log
   (2)}$$ which makes $x_1=1.44466736481181$ while the &quot;exact&quot; solution is
$x=1.44466736481170$.
Using one iteration of Halley method gives $21$ exact figures. Using one iteration of Householder method gives $27$ exact figures. The formulae are explicit but really too lengthy to be reported here.
Here is something very similar, and I think this YouTube video  may help. This is a video on how to solve equations of the type $$b^x =  x,$$ and the presenter solves for $x$ by taking the logarithm of both sides by doing:

$$ \ln(b^x) = \ln(x)$$
$$x \ln(b) = \ln(x)$$
$$x = \frac{ \ln(x)}{ \ln(b)}$$

In this case, the $b$ in the video is $1^{1/e}$. This problem seems very related - seeing as the solutions are the same, and the formulas seem similar when you substitute $x$ with $b$ and $2^x$ with $x$.