Setup: Suppose that $n \ {\rm exp}(-n\gamma)< \tau$ where $n \in \mathbb{N}, \tau \in (0,1], \gamma >0$. How to solve for $n$?
My attempt: Let $ n\triangleq {\rm ln} \ \alpha$, then we have the following:
$$ \implies {\rm ln} \ \alpha \ {\rm exp}(-{\rm ln} \ \alpha \ \gamma) < \tau$$
$$\implies {\rm ln} \ \alpha \ {\rm exp}(\ {\rm ln} \ (\alpha^{-\gamma})) < \tau$$
$$\implies {\rm ln} \ \alpha \times \frac{1}{(\alpha)^{\gamma}} < \tau$$
$$\implies \frac{\alpha -1}{\alpha} \frac{1}{(\alpha)^{\gamma}} < \tau$$
$$\implies (1 - \frac{1}{\alpha})\frac{1}{(\alpha)^{\gamma}} < \tau$$
where the second to last inequality follows from setting $1+x$ equals to $\alpha$, and using the inequality $$\frac{x}{1+x} \leq {\rm ln} \ x$$ when $x > -1$. Can someone help me with some suggestions on how to proceed further>
Want to solve $f(x) =xe^{-ax}=b, 0 < b \le 1, a > 0 $.
$f'(x) =e^{-ax}(1-ax) \gt 0$ for $0 < x < 1/a$, $=0$ for $x=1/a, \lt 0 $ for $x > 1/a$.
Therefore $f$ has a max at $x=1/a$ and $f(x)\to 0$ as $x \to \infty$.
At $x=1/a$, $f(x) =1/(ae)$ so no solution if $b > 1/(ae)$, one solution ($x=1/a$) if $b=1/(ae)$, and two solutions if $b < 1/(ae)$.
By writing the equation as $-axe^{-ax}=-ab $, the solutions can be written in terms of two branches of the Lambert W function as $\dfrac{W(-ab)}{-a} $.
For a numeric solution, if you don't have the Lanbert W function handy, start at $1/(2a)$ (for the smaller root) or $2/a$ (for the larger root) and apply Newton's iteration.
Playing a bit, here are the first Newton iterations starting at the suggestions above.
$g(x) =xe^{-ax}-b, g'(x) =e^{-ax}(1-ax), $.
$\begin{array}\\ g^*(x) &=x-\dfrac{g(x)}{g'(x)}\\ &=x-\dfrac{xe^{-ax}-b}{e^{-ax}(1-ax)}\\ &=x-\dfrac{x}{1-ax}+\dfrac{b}{e^{-ax}(1-ax)}\\ &=-\dfrac{ax^2}{1-ax}+\dfrac{b}{e^{-ax}(1-ax)}\\ &=\dfrac1{1-ax}(-ax^2+\dfrac{b}{e^{-ax}})\\ g^*(c/a) &=\dfrac1{1-c}(-c^2/a+\dfrac{b}{e^{-c}})\\ g^*(2/a) &=\dfrac1{-1}(-4/a+\dfrac{b}{e^{-2}})\\ &=4/a-e^2b\\ g^*(1/(2a)) &=\dfrac1{1/2}(-1/(4a)+\dfrac{b}{e^{-1/2}})\\ &=-1/(2a)+be^{1/2}\\ \end{array} $