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.

  • $\begingroup$ I think this one is out of the reach of even the Lambert W function, which does provide "closed forms" for things like $\ln x = 1/x.$ $\endgroup$ – coffeemath Nov 4 '14 at 2:51
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    $\begingroup$ You aren't going to find a nice closed form solution. $\endgroup$ – Thomas Andrews Nov 4 '14 at 2:51
  • $\begingroup$ Even Lambert won't help you with this one, I bet. @coffeemath $\endgroup$ – Thomas Andrews Nov 4 '14 at 2:51
  • $\begingroup$ If $e^x$ is near $4$, then $x\ge 1$ and $\ln x\ge 0$, so we know that both terms supply a positive contribution on the LHS. Suppose we try $1\le x\le \ln 4=2\ln 2\le 2$... Is there a series of approximations that gets us close? $\endgroup$ – abiessu Nov 4 '14 at 2:54
  • $\begingroup$ @ThomasAndrews Yes the OP equation won't get a W solution. However the simpler $\ln x=1/x$ does have one, as I recall. A fellow grad student once wrote it on the board, having forgotten to include $d/dx$ in the statement $d/dx \ln x=1/x,$ and we went to work on it, and found some solution using Lambert. [That was years ago, I may have misremembered it.] $\endgroup$ – coffeemath Nov 4 '14 at 3:28

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 !


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