Solving $\ln(x) = e^{-x}$

Question

I'm trying to solve $\ln(x) = e^{-x}$ but I can't really get how to do it :(

(Removing a statement that was incorrect, as explained by the comments below) Additionally, while I started to solve it I ended up with something really weird and I can't really understand what is the wrong passage:

Start with: $$ \ln(x) = e^{-x} $$

My understanding is that if $\ln(x) = a$ then that is equivalent to say that $e^{a} = x$ (assuming $x>0$), so, in the previous case:

$$ e^{e^{-x}} = x $$

And applying the exponent property that: $a^{m^{n}} = a^{mn}$

$$ e^{-ex} = x $$

But at this point it's really weird so I decided to take logarithm of both sides: $$ \ln(e^{-ex}) = \ln(x)\\ -ex\ln(e) = \ln(x)\\ -ex = ln(x)\\ $$

But that clashes with $ln(x) = e^{-x}$ because clearly $e^{-x}$ is not the same as $-ex$.

So could you please tell me which passage I'm doing wrong and how would you approach the initial equation?

Thanks a lot!

Answer 1

You won't find a closed formula for this equation. (Nor for the simpler $e^{-x}=x$, equivalent to $\ln(x)=-x$, though these can be expressed in terms of the Lambert W function, as $W(1)$.) You must resort to numerical root finding.

In this particular case, you will notice that the fixed-point method will work: start from any real value $x_0$ and iterate

$$x_{n+1}=e^{e^{-x_n}}.$$

This sequence always converges and at convergence, $x=e^{e^{-x}}$.

0
2.71828182846
1.06821393681
1.41004511415
1.27651316255
1.32181853101
1.30558334634
1.31129394458
1.30927185914
1.30998618969
1.30973363268
1.30982289996
1.30979134482
1.30980249885
1.30979855610
1.30979994979
1.30979945714
1.30979963128
1.30979956973
1.30979959149
...

Answer 2

Newton's Method says that we should iterate $$ x_{n+1}=\frac{x_ne^{-x_n}e^{e^{-x_n}}+e^{e^{-x_n}}}{e^{-x_n}e^{e^{-x_n}}+1} $$ This is because the right hand side is $$ x-\frac{f(x)}{f'(x)} $$ where $f(x)=e^{e^{-x}}-x$.

For example:

\begin{align} x_0&=0\\ x_1&=0.731058578630004879251159241822\\ x_2&=1.229791323145443105484685957301\\ x_3&=1.308681010785501028518648717394\\ x_4&=1.309799378172371994132312272864\\ x_5&=1.309799585804143328882655453650\\ x_6&=1.309799585804150477669233701960\\ x_7&=1.309799585804150477669233701968\\ x_8&=1.309799585804150477669233701968 \end{align}

Answer 3

by a numerical method i got this here $1.309799585804150477669233701968172506010868896430480435558475367426214513358226234915421428122420845$ you can use the Newton method