How to find a lower bound on a smallest solution of $\ln( \frac{x}{a} ) +\frac{1}{x}=0$. The equaiton
\begin{align}
 \ln( \frac{x}{a} ) +\frac{1}{x}=0
\end{align}
has two solutions for a sufficiently large $a$ (e.g., $a>3$).
I want a lower bound on the smaller of the two solutions.
I have found an upper bound on the largest of the two solutions as follows:
\begin{align}
0= \ln( \frac{x}{a} ) +\frac{1}{x}\le  1- \frac{a}{x}  +\frac{1}{x}
\end{align}
where we have used a lower boung $\ln(x) \ge 1-\frac{1}{x}$.  This implies that $x_2<a-1$.
However, I have no idea how to find a lower bound on the smallest solution.
 A: The function$$f(x)=\log \left(\frac{x}{a}\right)+\frac{1}{x}$$ has two solutions as soon as $a >e$.
As said in comments and answers, the smallest solution is given by
$$x=-\frac{1}{W_{-1}\left(-\frac{1}{a}\right)}$$
For the secondary branch, there exist nice bounds
$$-1 - \sqrt{2u} - u < W_{-1}\left(-e^{-(u+1)}\right) < -1 - \sqrt{2u} - \tfrac{2}{3}u$$ which make for your case
$$-\log (a)- \sqrt{2(\log (a)-1)}< W_{-1}\left(-\frac{1}{a}\right)$$
$$ W_{-1}\left(-\frac{1}{a}\right)< -\frac {2\log(a)+1}3 - \sqrt{2(\log (a)-1)}$$
For $a=e^2$,the smallest solution is then bounded by
$$1-\frac{1}{\sqrt{2}} \sim 0.292893 < x <\frac{3}{7} \left(5-3 \sqrt{2}\right)\sim 0.324583$$ while the exact solution is $\sim 0.317844$.
So, a first lower bound is
$$\color{red}{x_0=\frac 1 {A+ \sqrt{2(A-1)} }}\quad \text{where}\qquad \color{red}{A=\log(a)}$$
However, this can be improved a lot since $f(x_0) >0$ and $f''(x_0)>0$. So, by Darboux theorem, the solution will be reached without any overshoot of the solution and better bounds will be given by
$$\color{red}{x_{n+1}=x_n+\frac{x_n \left(x_n\log \left(\frac{x_n}{a}\right)+1\right)}{1-x_n}}$$ For example
$$\color{red}{x_1=\frac{\left(A+2  \sqrt{2(A-1)}-1\right)-\log \left(A+
   \sqrt{2(A-1)}\right)}{\left(A+ \sqrt{2(A-1)}-1\right) \left(A+   \sqrt{2(A-1)}\right)}}$$
For the working case $(a=e^2)$, the left bounds are sucessively
$(n=0,1,2)$
$$\{0.2928932188,0.3154911206,0.3178230036\}$$
A: Set $u = \ln(x/a)$. Then this gives
$$
u + \frac{1}{ae^u} = 0
$$
which we rewrite as
$$
u e^u = -\frac{1}{a}
$$
Then we see
$$
u = W(-1/a)
$$
Where $W$ is the Lambert W function.
Now it's known (it's actually on the wikipedia page linked above) that
$$
W(x) = \sum_{n = 1}^\infty \frac{(-n)^{n-1}}{n!} x^n
$$
whenever $|x| < 1/e$.
Thankfully you added the requirement that $a > 3$ or so, and so we can use this formula to derive progressively better lower bounds, since we know the exact solution will be
$$
\ln(x/a) = u = W(-1/a) = 
\sum_{n = 1}^\infty \frac{(-n)^{n-1}}{n!} \left ( \frac{-1}{a} \right )^n
$$

I hope this helps ^_^
