A certain generalisation of the golden ratio 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
 A: Here is a way to improve your lower bound (l.b.) and upper bound for large enough $a$. Let $x_a:=\text{lng}(a)$. Then for $x=x_a$ we have 
\begin{equation}
 1-x+e^{-ax}<e^{-x}+e^{-ax}=1,
\end{equation}
whence $x>e^{-ax}$ and $ax e^{ax}>a=aye^{ay}$, where 
\begin{equation}
 y=y_a=\frac{L(a)}a
\end{equation}
and $L$ is the Lambert function, so that $L(a)e^{L(a)}=a$ and $L(a)>0$. So, $y_a=\frac{L(a)}a$ is a l.b. on $x_a$. One can also use the elementary l.b. 
\begin{equation}
 L_{\text{low}}(a):=\ln a - \ln\ln a + (\ln\ln a)/(2 \ln a)
\end{equation}
on $L(a)$ for $a>e$, 
found near the end of Section "Asymptotic expansions" at Lambert function, to we obtain an elementary l.b., say $\tilde y_a$, on $y_a$. 
Similarly, for $x=x_a$, using the already proved inequality $x_a>y_a$, we have 
\begin{equation}
 \frac1{1+y_a}+e^{-ax}>\frac1{1+x}+e^{-ax}>e^{-x}+e^{-ax}=1,
\end{equation}
whence $e^{-ax}>1-\frac1{1+y_a}$ and
\begin{equation}
 x=x_a<z_a:=\frac1a\,\ln\Big(\frac1{y_a}+1\Big), 
\end{equation}
so that $z_a$ is an upper bound on $x_a$. Moreover, replacing $y_a=\frac{L(a)}a$ in the above expression for $z_a$ by its lower bound $\frac{L_{\text{low}}(a)}a$, we obtain an elementary u.b., say $\tilde z_a$, on $z_a$. 
Thus, 
\begin{equation}
 \tilde y_a<y_a<x_a<z_a<\tilde z_a. 
\end{equation}
For $a\to\infty$, we have $L(a)\sim \ln a$ and hence $y_a\sim \frac{\ln a}a$ and $z_a\sim y_a$, so that the l.b. $y_a$ and the u.b. $z_a$ on $x_a$ are both asymptotically optimal. Moreover, since $L(a)\sim L_{\text{low}}(a)$, the elementary l.b. $\tilde y_a$ and u.b. $\tilde z_a$ on $x_a$ are also asymptotically optimal. 
Your l.b. and u.b. seem to be good for small values of $a$, whereas the l.b. and u.b. on $x_a$ obtained here are good for large enough $a$. Here is an illustration: 

