Can the "simple" equation $e^x=\log(x)$ be solved using algebra?

Question

I came across this really simple-looking yet astonishingly hard problem to solve: $$e^x=\log(x).$$

I tried to use Lambert-W function, but I cannot get it to the required standard form. Even Wolfram Alpha just provides the numerical approximation. Is this solvable using any special functions?

Answer 1

Not algebraic, but maybe better than nothing. Assuming $\log=\ln$: In $$ e^x=\ln x \tag1 $$ multiply both sides with $x$: $$ xe^x=x\ln x \tag2 $$ and then apply Lambert-W on both sides using$^1$ $W(xe^x) = x$ and $W(x\ln x)=\ln x$:

$$ x=\ln x \tag 3 $$ Now $\ln (1/x) = -\ln x$ so we can divide by $x$ to get:

$$ -1=\frac1x\ln\frac1x \tag 4 $$ and then use $W(x\ln x)=\ln x$ again:

$$ W(-1)=W\left(\frac1x\ln\frac1x\right)=\ln\frac1x=-\ln x \tag 5 $$ hence $$ x = \exp (-W(-1)) \tag 6 $$ Wikipedia mentions the approximated $W_0(-1) \approx -0.31813+1.33723i$.

Now this is a bit sloppy w.r.t. to which branch is being considered, but it should be clear now how $(1)$ can be approached.

---

$^1$See for example Wikipedia

Answer 2

For $x>0$, the equation doesn't have solutions because $e^x>\ln(x)$ for all $x>0$, as e.g. the plot at Wolfram Alpha shows.

$$e^x=\ln(x)$$

We see, your equation is an irreducible polynomial equation of more than one algebraically independent monomials ($e^x,\ln(x)$). We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.
I don't know if the equation has solutions in the elementary numbers.

$x\to e^t$: $$e^{e^t}=\ln(e^t)$$ $$e^{e^t}=t+2k\pi i\ \ \ (\forall k\in\mathbb{Z})$$

We see, we cannot solve the equation with Lambert W in the general case, but with Hyper Lambert W:

$$1=e^{-e^t}(t+2k\pi i)$$ $$e^{-e^t}(t+2k\pi i)=1$$ $t\to u-2k\pi i$: $$ue^{-e^{u-2k\pi i}}=1$$ $$ue^{-e^{-2k\pi i}e^u}=1$$ $$G(-e^{-2k\pi i};u)=1$$ $$u=HW(-e^{-2k\pi i};1)$$ $$t=HW(-e^{-2k\pi i};1)-2k\pi i$$ $$x=e^{HW(-e^{-2k\pi i};1)-2k\pi i}$$

So we have a closed form for $x$, and the representations of Hyper Lambert W give some hints for calculating $x$.

Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997

Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119

Answer 3

We have $e^{e^x}=x$. Multiply by $e^x$ to get $e^x e^{e^x}=x\ e^x$. Apply $W$ to get $e^x =x$, so $-1=-x\ e^{-x}$ and $x=-W(-1)$. This only gives one solution because we are using the relation $W(z e^z)=z$ which is only valid in a certain region of $\Bbb C$.