Finding $\log_{-e} e$

Question

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?

Answer 1

$$x=\log_{-e}e \implies (-e)^x=e \\ e^{1/x}=-e \\ \frac{e^{1/x}}e=-1 \\ e^{1/x-1}=-1$$

Now, take the identity that $e^{i\pi}=-1$

$$e^{1/x-1}=e^{i\pi} \\ \frac1x-1=i\pi \\ \frac1x = i\pi+1 \\~\\ x=\frac1{i\pi+1}$$

Answer 2

$$\log_ab=\frac{\log b}{\log a}$$

$$\log (-e)=1+i\pi$$

Answer 3

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

$$\frac{1}{1+i\pi} \cdot \frac{1-i\pi}{1-i\pi} = \frac{1-i\pi}{1 + \pi^2}$$

Answer 4

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. $$t = \dfrac{1+2mi\pi}{1+(2n+1)i\pi}$$