algebra-precalculusclosed-formexponential functionlogarithms
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?
There is no known closed-form solution for equations like this one. Let's simplify it: $$\log^2x^2+\log(x-1)=0$$ $$\log x^2\cdot\log x^2=-\log(x-1)$$ $$\log x^{2\log x^2}=\log\dfrac1{x-1}$$ $$x^{4\log x}=\dfrac1{x-1}$$ $$x^{-4\log x}=x-1$$ We have exponential and logarithmic function on LHS and linear function on RHS. When we have linear and exponential or logarithmic function in same equation we must use Lambert W function. Only equation which can be solved using Lambert W function is $$a^{f(x)}+bf(x)+c=0$$ if we can find $f^{-1}(x)$ where $a,b,c$ are constants. Equation $\log^2x^2+\log(x-1)=0$ cannot be reduced to this form, so there are no closed-form solution.
More than a full answer, this is a first step in looking for a solution.
Consider a function $f(k)$ such that $$ \Delta \,f(k) = \Delta \,x = \left( \matrix{ n \cr k \cr} \right)p^{\,k} \left( {1 - p} \right)^{\,n - k} $$
That is precisely the Cumulative of the binomial distribution you give $$ C_X (k\,;\,n,p) = \sum\limits_{j = 0}^{k - 1} {\binom{n}{j} p^{\,j} \left( {1 - p} \right)^{\,n - j} } $$
We shall make it continuous in $k$, using
for the sum the Indefinite Sum, i.e.
$$
\eqalign{
& \Delta F(x) = F(x + 1) - F(x) = f(x)\quad \Rightarrow \cr
& \Rightarrow \quad F(x) = \Delta ^{\, - 1} f(x) = \sum\nolimits_{\;j = 0}^{\;x} {f(j)} \cr}
$$
So we write $$ C_X (k\,;\,n,p) = \left( {1 - p} \right)^{\,n} \sum\nolimits_{\;j = 0}^{\;k} {\binom{n}{j}\left( {{p \over {1 - p}}} \right)^{\,j} } $$
The floor of its functional inverse (in $k$) is then the function you are looking for.
But unfortunately there is no close expression available for the partial sum on the lower index of a binomial.
Best Answer
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