Question: I have a recurrence relation, it is like the following:

$T(e^n) = 2(T(e^{n-1})) + e^n$, where $e$ is the base of the natural logarithm.

To solve this and find a $\Theta$ bound, I tried the following: I put $k=e^n$, and the equation transforms into:

$$T(k)=2T(k/e)+k$$

Then, I try to use the Master Theorem. According to Master Theorem, $a=2$, $b=e\gt 2$ and $f(k)=k$. So, we have the case where $f(k)=\Omega(n^{\log\_b a+ε})$ for some $ε\gt 0$, thus we have $T(k)=\Theta(f(k))=\Theta(k)$. Then put $k=n$, we have $T(n)=\Theta(n)$. Does my solution have any mistakes?

Answer: 
$c=1\gt \ln 2\approx0.7$. That's [case 3](http://en.wikipedia.org/wiki/Master_theorem) of the master theorem. It follows that:

$$T(n)=\Theta(n)$$

Your solution is fine.
Just use the change of variables $t(n) = T(e^n)$, you get $t(n) = t(n - 1) + e^n$. Linear, first order recurrence, solution is:

$\begin{equation\*}
 t(n)
 = t(n - 1) + e^n
\end{equation\*}$

Solution is:

$\begin{align\*}
 t(n)
 &= t\_0 - \frac{e}{e - 1} + \frac{e^{n + 1}}{e - 1} \\
 T(k)
 &= (t\_0 - \frac{e}{e - 1}) + \frac{e^2}{e - 1} k \\
 &\sim \frac{e^2}{e - 1} k
\end{align\*}$