Finding $\log_{-e} e$ I'm wondering if it is possible to solve for a negative base logarithm, $-e$ for $e$. From Euler's formula, we can find the logarithm of complex numbers in an instrumental fashion. However, would a negative base of $e$ as a logarithm still make sense? 
 A: $$x=\log_{-e}e \implies
(-e)^x=e \\
e^{1/x}=-e \\
\frac{e^{1/x}}e=-1 \\
e^{1/x-1}=-1$$
Now, take the identity that $e^{i\pi}=-1$
$$e^{1/x-1}=e^{i\pi} \\
\frac1x-1=i\pi \\
\frac1x = i\pi+1 \\~\\
x=\frac1{i\pi+1}$$
A: $$\log_ab=\frac{\log b}{\log a}$$
$$\log (-e)=1+i\pi$$
A: Your log is in base $-e$. You can always change basis according to the law:
$$\log_b(x) = \frac{\log_d\ x}{\log_d\ b}$$
Where $d$ is your new basis.
Let's choose a new basis like $+e$ and you get
$$\log_{-e} e = \frac{\log_e e}{\log_e (-e)}$$
Now $\log_e e = 1$ and thanks to complex analysis you have 
$$\log_e(-e) = \log_e(e\cdot -1) = \log_e(e) + \log_e(-1) = 1 + i\pi$$
Hence
$$\log_{-e} e = \frac{1}{1 + i\pi}$$
Eventually you can rationalise and find
$$\frac{1}{1+i\pi} \cdot \frac{1-i\pi}{1-i\pi} = \frac{1-i\pi}{1 + \pi^2}$$
A: By definition, $\log_b(z)$ is any complex number $t$ such that $b^t = z$.  But $b^t$ is defined as $\exp(t \log b)$, where $\log b$ is any branch of the natural logarithm of $b$, i.e. any complex number $s$ such that $\exp(s) = b$.
In your case you have $\log (-e) = 1 + (2n+1)i\pi$ for arbitrary integer $n$, and
$$\exp(t (1+(2n+1)i\pi)) = e = \exp(1)$$ iff for some integer $m$,
$$ t(1+(2n+1)i\pi) = 1 + 2mi\pi$$
i.e. $$t = \dfrac{1+2mi\pi}{1+(2n+1)i\pi}$$
