Consider a real number $a \ge 1$, and let $g(a)$ be the unique positive solution $x>0$ of $x^a - x^{a-1} - 1 = 0.$ We have $g(1) = 2$, $g(2) = {1+\sqrt{5}\over2}$ (the golden ratio), and $g$ is strictly decreasing, with limit $1$. As bounds we know $$4^{1\over 1+a} \le g(a) \le 2^{a^{-1/2}}.$$

Of interest is also $\operatorname{lng}(a) := \ln(g(a))$ (using the natural logarithm). So $\operatorname{lng}(1) = \ln(2)$, and $${\ln(4)\over 1+a} \le \operatorname{lng}(a) \le \ln(2) a^{-1/2};$$ $\operatorname{lng}(a)$ is the unique positive solution $x > 0$ of $\exp(-x) + \exp(-ax) = 1.$

Any information on $g(a)$ (or $\operatorname{lng}(a)$) would be helpful. We are especially interested in good bounds, especially in good lower bounds for $g(a)$ ($\operatorname{lng}(a)$).

Restricting parameter a to natural numbers is also natural here. For $a$ in $\{1,2,3,4\}$ there are solutions via radicals for $g(a)$, while for other values of a we are not aware of any. Information on this would also be interesting.

The background can be seen in the study of recurrences like $f(n) = f(n-1) + f(n-5)$, belonging to g(5). For that quantities like $g(a)$ need to be approximated well.

The natural way of computing $\operatorname{lng}(a)$ is to use Newton approximation, starting with the lower bound ${\ln(4)\over 1+a}$. To reach (close to) full precision with a floating point type needs in case of "long double" (an extended $80$ big floating point type) quite precisely $\ln(a)$ many iterations. Having a better lower bound $\operatorname{lb}(a)$ for $\operatorname{lng}(a)$, we could start the computation with $\operatorname{lb}(a)$ instead, using fewer iterations then.

I have been using implicitly $g(a)$ in my work for many years now, but never came across some systematic study of such a function.

ANSWER to comments and Answer: Thank you for your comments, all very useful. The general lower bound by Iosif Pinelis should be very helpful. Unfortunately I try now already for quite some time all possibilities to add a comment, but it all fails, only this edit-function works. So sorry for this misuse. Oliver