Exact, closed-form solution for $e^x + ln(x) = 4$? I have been working on this problem that I have heard was posted in a high-school pre-calculus course. As I would expect, there should be a closed-form expression for $e^x + ln(x) = 4$ for some $x$, but WolframAlpha does not give one, only an approximate numerical answer.
Is there a way of solving this without the use of calculus or any other higher-order methods?
I have been able to get it to $xe^{e^x} = e^4$, but that does not seem to help anything.
 A: If the problem was to find the solution of $$e^x+x=4$$ the solution would have been $$x=4-W\left(e^4\right)$$
If the problem was to find the solution of $$x+\log(x)=4$$ the solution would have been $$x=W\left(e^4\right)$$
But the problem is to find the solution of $$e^x + ln(x) = 4$$ for which it does not seem to have an analytical solution. However, solving numerically shows that the solution is $x\approx 1.31531$ and looking at $RIES$, this number is very close to $$x_0=\frac{\sqrt{e+\frac{1}{\sqrt{e}}}}{ \log (4)} $$
To polish the solution, let us use one single iteration of Newton and get $$x_1=\frac{\sqrt{\frac{1}{\sqrt{e}}+e}}{\log
   (4)}-\frac{-4+e^{\frac{\sqrt{\frac{1}{\sqrt{e}}+e}}{\log (4)}}+\frac{1}{2} \log
   \left(\frac{1}{\sqrt{e}}+e\right)-\log (\log
   (4))}{e^{\frac{\sqrt{\frac{1}{\sqrt{e}}+e}}{\log (4)}}+\frac{\log
   (4)}{\sqrt{\frac{1}{\sqrt{e}}+e}}}$$
It should be notice that $f(x_1)=3.711\times 10^{-11}$ while $f(x_0)=-2.179\times 10^{-5}$
Not the exact solution but not too far !
