Unforeseen issue in my MastersThesis: Is there a "closed form" solution? I'm an electrical engineer and I recently came across an unforeseen issue in my masters thesis because I lack a deeper mathematical education.
I want to know for which positive real $x$ the following inequality is an equality:
$$
n \log(1 + \tfrac{x}{n}) - \log(1 + x) \leq a\, , \;\;\;\; (\ast)
$$
i.e.
$$
n \log(1 + \tfrac{x}{n}) - \log(1 + x) - a = 0\, ,
$$
where $x \in \mathbb{R} \geq 0$, $n\in \mathbb{N}\gg 1$ and $a \in \mathbb{R} > 0$.
This is equivalent to
$$
(1+\tfrac{x}{n})^n = \mathrm{e}^a\cdot(1 + x) \,
$$
which looks basically not so hard.
My questions is:
Is there a "closed form" solution of this equation for the unknown $x$ and I am just too dumb to get it?
By closed form solution I mean any solution that I can nicely write like 
$$
x \leq \ldots
$$
to solve $(\ast)$.
If there is no "closed form" solution, I would be interested why and how I could have seen this. Unfortunately I'm not really familiar enough with Transcendence Theory or Galois Theory to see this on my own.
Thanks!

Assuming that $(\ast)$ has no "closed form" solution, I also thought about a workaround.
Since $\lim\limits_{n \rightarrow \infty} (1 + \tfrac{x}{n})^n = \mathrm{e}^x$, I could write write
$$
\mathrm{e}^x - \mathrm{e}^a\cdot(1 + x) = 0 \, 
$$
for the case that $n \rightarrow \infty$.
I thought about this as an approximation for finite $n$.
Unfortunately for finite $n$, $(1 + \tfrac{x}{n})^n < \mathrm{e}^x$ yields and this approximation would clash my basic inequality $(\ast)$.
If I could show that
$$
\lim\limits_{n \rightarrow \infty} (1 + \tfrac{x}{n})^{n+2} = \mathrm{e}^x
$$
and
$$
(1 + \tfrac{x}{n})^{n+2} > \mathrm{e}^x
$$
for finite $n$, it might be possible to solve 
$$
\mathrm{e}^x\cdot(1 + \tfrac{x}{n})^{-2} =  \mathrm{e}^a\cdot(1 + x)
$$
for an approximate solution of $(\ast)$.
Here, the Lambert W function might be helpful but I didn't succeed on this problem so far as well.
Any thoughts on this?
Thanks a lot!
 A: Consider the equation $\left(1+\frac{x}{n}\right)^n=e^a(1+x)$.
Choose $c\in\mathbb{C}$ such that $c^{n-1} = -\frac{1}{n}e^{-a}$ and define $y = c\left(1+\frac{x}{n}\right)$, then the equation becomes
$$c^{-n}y^n = e^a\left(\frac{n}{c}y+1-n\right)$$ and after multiplying with $c^n$
$$ y^n = e^a\left(nc^{n-1} y+(1-n)c^n\right).$$
Using $c^n = cc^{n-1} = -\frac{c}{n}e^{-a}$ this may finally be rewritten to
$$ y^n +y= c\left(1-\frac{1}{n}\right).$$
The solution of this equation can in general not be written in terms of roots and elementary arithmetic operations, but it can in terms of an ultraradical:
$$y = \mathrm{ur}\left(c\left(1-\frac{1}{n}\right)\right)$$
and hence
$$x = \frac{n}{c}\mathrm{ur}\left(c\left(1-\frac{1}{n}\right)\right)-n.$$
A: I think you should continue like this ($\ t=1+x$):
$$
e^x=e^a(1+x)\\
\frac{e^x}{(1+x)}=e^a\\
-te^{-t}=-e^{-(a+1)}\\
W(-e^{-(a+1)})=-t\\
t=-W(-e^{-(a+1)})\\
x=1-W(-e^{-(a+1)})
$$
A: As noted, your equation is the same as $$\left(1+\frac{x}{n}\right)^n=e^a(1+x)$$ Take a look at the graphs of $y=\left(1+\frac{x}{n}\right)^n$ and $y=e^a(1+x)$. The first is a polynomial with a single negative repeated root. The other is a line with slope $e^a$ and root at $-1$. When $n$ is even, there are two solutions, one positive and one negative (but larger than $-1$). If $n$ is odd, there is still a positive solution, but a negative solution only exists if $a$ is large enough to give that line steep slope.
For $n$ larger than 4, there is no general solution by radicals for such a polynomial equation. I would recommend arguing that solutions exist as I have done (only more formally) and noting that when needed, they can be found to arbitrarily high precision using any one of a number of methods, including Newton's method.
