Root estimation What is the estimation for the positive root of the following equation
$$
ax^k = (x+1)^{k-1}
$$
where $a > 0$ (specifically $0 < a \leq 1$). 
Could you point out some reference related to the question?
 A: Put $\displaystyle z = 1 + \frac{1}{x}$ and we get the equation
$$z^{n-1}(z-1) = a$$
(I prefer to use $\displaystyle n$ instead of $\displaystyle k$)
We can easily see that $\displaystyle z \in (1,2)$
Assume that $\displaystyle z = 1 + \frac{g(n)}{n-1}$
Thus we have that
$$\left(1 + \frac{g(n)}{n-1}\right)^{n-1} g(n) = (n-1)a$$
Now we have that $\displaystyle e^{x/2} \lt 1 + x \lt e^x$ for $\displaystyle x \in (0,1)$
And so we get $$e^{g(n)} g(n) \gt (n-1)a \gt e^{g(n)/2} g(n)$$
Now since $\displaystyle xe^x$ is increasing, and the root of $\displaystyle xe^x = y$ is given by the LambertW function: $\displaystyle W(y)$.
Thus we get that
$\displaystyle g(n) \gt W((n-1)a)$
and
$\displaystyle g(n) \lt 2 W\left(\frac{(n-1)a}{2}\right)$
It is well known that $\displaystyle W(x) = \theta(\log x)$ (as $\displaystyle x \to \infty$) and so we get that
$\displaystyle g(n) = \theta (\log (n-1)a)  = \theta(\log na)$
Thus the root $\displaystyle r(n)$ is $$\theta\left(\frac{n}{\log na}\right)$$
In fact, I would go so far as to guess that
$$ \lim_{n \to \infty} \dfrac{r(n)W((n-1)a)}{n-1} = 1$$
