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

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!

• What? In general, $a^{m^n}\ne (a^m)^n=a^{mn}.$
– mfl
Commented Oct 27, 2014 at 20:05
• "And applying the exponent property that: $a^{m^{n}} = a^{mn}$" <- No, that's wrong. $(a^m)^n = a^{mn}$, not $a^{m^n}$, which is $a^{(m^n)}$. Commented Oct 27, 2014 at 20:05
• Argh sorry you are right :| Commented Oct 27, 2014 at 20:06

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
...


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}

• I wanted the table on the left. Did it seem that off-balance?
– robjohn
Commented Oct 27, 2014 at 21:12

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

• how do you get to that number? indeed it is the right result Commented Oct 27, 2014 at 20:09
• @complexguest: I don't know how the author did it, but this is the answer that Mathematica returns for FindRoot[Exp[Exp[-x]]-x,{x,1},WorkingPrecision->100]
– robjohn
Commented Oct 27, 2014 at 21:20