Root of $(x+a)^{x+a}=x^{x+2a}$ and $e$ 
Let us denote solution to the equation
$$(x+a)^{x+a}=x^{x+2a}$$
with $X_a$.
($a$ is a non-zero real number)
Prove that:
$$\lim_ {a \to 0} X_a = e$$

This is something that I noticed while making numerical experiments for another problem. The statement looks interesting, I couldn't find anything close to it on the internet. I don't have the idea how to prove it, but numerical methods confirm the statement.
 A: This might not be rigorous, but note that one has 
$$(x+a)^x=x^{x+a}\Rightarrow x\ln(x+a)=(x+a)\ln x\Rightarrow \frac{\ln(x+a)}{x+a}=\frac{\ln x}{x}.$$
Then, let $f(x)=\frac{\ln x}{x}$. One has $f'(x)=0\iff x=e$, and considering the graph of $y=f(x)$ should give you the answer.
A: Taking the logarithm of both sides of $$(x+a)^{x+a}=x^{x+2a}$$
we get 
$$(x+a)\ln(x+a)=x\ln(x)+2a\ln(x)$$
or
$$\frac{(x+a)\ln(x+a)-x\ln(x)}{a}=2\ln(x). \tag 1$$ 
The left hand side tends to $\frac{d(x\ln(x))}{dx}=\ln(x)+1$ if $a$ tends to zero. The right hand side does not depend on $a$. That is,
$$\ln(x)+1=2\ln(x).$$
As a result, whatever the solution of the original equation is, it has to tend to $e$ if $a$ tends to zero.
EDITED
To be precise, let's start over again from $(1)$. First of all, make it clear that $x$ is a solution of $(1)$ and, as a result, it depends on $a$. Also Let's  subtract $\ln(x_a)+1$ from both sides of $(1)$. Having taken the absolute value of both sides, we have now:
$$\left|\frac{(x_a+a)\ln(x_a+a)-x_a\ln(x_a)}{a}-(\ln(x_a)+1)\right|=|\ln(x_a)-1|.$$
For any $\delta>0$ and $\epsilon>0$ there exists an $a(\delta,\epsilon)$ such that $\epsilon>a(\delta,\epsilon)>0$ and 
$$|\ln(x_{a(\delta,\epsilon)})-1|<\delta.\tag 2$$
Now, assume that $$\lim\limits_{a\rightarrow 0}x_{a}=e+\xi,\tag 3$$ for a $|\xi|>0$. Since $\ln(x)$ is continuous at $e$, $(2)$ and $(3)$ are contradicting.
A: Rewriting the equation to $(1+\frac{a}{x})^{x+a}=x^a$ and taking $\ln$, we have $(x+a) \ln(1+\frac{a}{x})=a\ln x$. Assuming that $X_a$ is bounded for $a$ from some neighbrhood of $0$, we take the Taylor approximation for small $a$ of the left hand side:
$$
(X_a+a) (\frac{a}{X_a}+o(a))=a\ln X_a
$$
and
$$ 
a+o(a) = a\ln X_a
$$
which yields $\ln X_a=1$.
A: If $a$ is small, a first order Taylor expansion built at $a=0$ gives $$(x+a)^{x+a}-x^{x+2a}=-a\, x^x\, \big(\log (x)-1\big)+O\left(a^2\right)$$ Do you think that this is sufficient to prove that $$\lim_ {a \to 0} X_a = e$$
Another way is starting from zoli's answer $$\frac{(x+a)\log(x+a)-x\log(x)}{a}=2\log(x)$$ and expand the lhs as a Taylor series built at $a=0$ gives $$(\log (x)+1)+\frac{a}{2 x}+O\left(a^2\right)\approx 2\log(x)$$ that is to say $$\frac{a}{2 x}-\log (x)+1\approx 0$$ and the solution of what would be the equation is $$x\approx \frac{a}{2 W\left(\frac{a}{2 e}\right)}=e+\frac{a}{2}-\frac{a^2}{8 e}+O\left(a^3\right)$$ where appears once more Lambert function.
Edit
What could be amazing when knowing the answer is to write $$x=e+\sum_{i=1}^\infty \alpha_i a^i$$ and to expand the equation around $a=0$. For the first terms, we have $$x=e+\frac{a}{2}-\frac{7 a^2}{24 e}+\frac{a^3}{3 e^2}+O\left(a^4\right)$$ For example, for $a=1$ this approximation gives $x\approx 3.15610$ while the rigourous solution should be  $x\approx 3.14104$ 
