According to Google, $\log(-1) = 1.36 i$. Why is that true? Euler's identity says $e^{i\pi} = -1$. Taking the log on both sides gives $\log(-1) = i\pi (\neq 1.36i)$.
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4There might be some confusion with your notation of $\log$ since that usually refers to base 10. Try using $\ln$ – aleden Jan 02 '18 at 15:40
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1You should write $ln$ instead of $log$ on Google. – Zubzub Jan 02 '18 at 15:41
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$\log$ is generally taken to base ten. Try $\ln$ which is to base $e$ . And you have to use $\ln$ to get $\pi i$ in L.H.S – user167920 Jan 05 '18 at 03:43
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Because it interpreted $\log$ as the base 10 logarithm.
For the natural logarithm $\ln$ it truly is :
$$e^{i\pi}=-1 \Rightarrow \ln(e^{i\pi})=\ln(-1)\Rightarrow \ln(-1)=i\pi$$
For the base 10 logarithm, it is :
$$e^{i\pi}=1\Rightarrow \log_{10}(e^{i\pi})=\log_{10}(-1)\Rightarrow\log_{10}(-1)=i\pi\log_{10}(e)\approx1.36i$$
Rebellos
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In addition to what the other answers have told you about conflating $\ln$ with $\log_{10}$, it is also worth noting that there is no "canonical" choice of a complex logarithm function; one could also consistently say that, e.g., $\ln(-1) = 3\pi i$ since $e^{3\pi i} = e^{2\pi i + \pi i} = e^{2\pi i}e^{\pi i} = 1 e^{\pi i} = -1$. See also: https://en.wikipedia.org/wiki/Complex_logarithm
Gregory J. Puleo
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Starting from $e^{i\pi} = -1$, we have $\log(-1) = i\pi \log(e)$.
What is $\log(e)$? It depends on who you ask. Many mathematicians use log as an abbreviation for the natural logarithm (base $e$), while others (including most calculators) use it for the common logarithm. Indeed $$ \pi \log_{10}(e) \approx 1.364 $$ so I suspect that's why Google tells you what it does.
Matthew Leingang
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