Why is $\log(-1) = 1.36 i$ instead of $\pi i$? 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 $\log(-1) = i\pi (\neq 1.36i)$.
 A: 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 : 
$$e^{i\pi}=1\Rightarrow \log_{10}(e^{i\pi})=\log_{10}(-1)\Rightarrow\log_{10}(-1)=i\pi\log_{10}(e)\approx1.36i$$
A: 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
A: 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.
