To solve $\displaystyle 2^x = x^5 $ could you use Lambert W function?
Thanks
In principle yes it can, the equation can be transformormrd to:
$\displaystyle
\frac{-x \ln(2)}{5}e^{\frac{-x \ln(2)}{5}}=\frac{- \ln(2)}{5}
$
So formaly we have:
$\displaystyle
\frac{-x \ln(2)}{5} = W\left(\frac{- \ln(2)}{5}\right) \approx W(-0.1386)
$
Now the problem is that there a multiple (real) branches of the Lambert W in
the region around $\displaystyle -0.1386$, and you are going to have to
evaluate two of them to get all the solution to this. My LambertW calculator
gives $\displaystyle x\approx 1.17724$.
However just plot the functions and you will see that there is another
solution near $\displaystyle x=22$ Numerical methods will find both solutions
once you know where they are.
RonL