What is a logarithm of a number with a negative base?

Question

Say I have $$\log_{-2}x$$How would I evaluate this? I am not against complex results or formulas, I just am curious.

Edit: There must be an answer, as $$\log_{-2}4=2$$

Answer 1

Let $$\log_{-2}x = c$$

Then $$(-2)^{c} = x$$

$$-2 = x^{\frac{1}{c}}$$

Note that $e^{i \pi} = -1$

$$\implies 2e^{i\pi} = x^{\frac{1}{c}}$$

$$x = 2^{c}e^{c\cdot i\pi }$$

So in your example, if we are given $c=2$, then $x = 2^2\cdot e^{2i\pi}= 4(\cos(2\pi) + i\sin(2\pi)) = 4$

Answer 2

if $a^x=b$, then $\log_a(b) = x$. Moreover, $\log_a(b) = \frac{\ln(b)}{\ln(a)}$

Let's set $a=-2$ and see what happens.

$\log_{-2}(x) = \frac{\ln(x)}{\ln(-2)}$

We know that $\ln(-2) = k$ where $e^{k}=-2$. As $e^{i\pi} = -1$, it follows that $2e^{i\pi} = -2$. Hence $\ln(-2) = \ln(2e^{i\pi}) = \ln(2)+i\pi$.

Now you have a formula for $\log_{-2}(x)$: \begin{align*} \log_{-2}(x) = \frac{\ln(x)}{\ln(2)+i\pi} \end{align*}

There's a problem with this though, because for any real $x$ (e.g. $x = 4$) you will always have a complex logarithm. This clearly contradicts $\log_{-2}(4) = 2$. There must be something subtle hidden in the fact that I chose $\ln(-2) = \ln(2)+i\pi$, when in reality $\ln(-2) = \ln(2)+ni\pi$ for every odd integer $n$.