# Is $\log_{5}{-3} = \frac{\log(3)+\pi i}{\log(5)}$?

Why does my calculator return false when I input $\log_{5}{-3} = \frac{\log(3)+\pi i}{\log(5)}$ but W|A returns true?

I'm thinking my calculator is wrong because I know that $\displaystyle \log_{5}{(-3)} = \frac{\log(-3)}{\log(5)}$ and $\displaystyle \log(-x) = \log(x) + \pi i$ so that means that $\displaystyle \log_{5}{(-3)} =\frac{\log(3) + \pi i}{\log(5)}$.

So why does my calculator return false?

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What does your calculator say for that logarithm? False? Nonexistent? Maybe your calculator uses real numbers and considers logs of negative numbers nonexistent .... what does your calculator say for square-root of -3 ????? –  GEdgar Jun 24 '12 at 17:06
No. I have a special application my calculator through which I am able to make logical expressions. Like 5 = 5 will return true on that app. My calculator is giving me false for this logic. –  David Jun 24 '12 at 20:18

Complex logarithms are multivalued, since the exponential is periodic with period $2 \pi i$. Thus, properly speaking, $$\log_5(-3) = \frac{\log(3) + \pi i + 2 n \pi i}{\log(5) + 2 k \pi i} n,k \in \mathbb{Z}$$
So, your calculator may be using different values of $n$ and $k$ from your $n=k=0$. In the vocabulary of complex analysis, we might say that your calculator is returning a value from a different branch than you expect.
Sorry. I had made I mistake. I was comparing my calculator with WA's calculator and it slipped by mind that default logarithm WA uses is the natural logarithm. I was using the log with a base of 10 instead of with a base of $e$ –  David Aug 3 '12 at 17:29