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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 $$\log_{-2}4=2$$

Michael Hardy
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    You might find this thread interesting. – Théophile Mar 26 '17 at 01:41
  • There must be an answer as log base neg 2 is 4. That only means there is sometimes a solution. if you limit your self to real numbers the only possible values are rational numbers and the only possible input are those values that are rational powers of that particular base. lob base neg of pos 8 has to real solution. Complex values are an entirely different story. – fleablood Mar 26 '17 at 04:25

2 Answers2

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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 $n$.

Harambe
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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 $x = 2^2\cdot e^{2i\pi}= 4(\cos(2\pi) + i\sin(2\pi)) = 4$

mrnovice
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