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Question Number 91936 by MWSuSon last updated on 03/May/20

$$howtoevaluateln\left(i\right),i=\sqrt{-1}.$$

Commented by MWSuSon last updated on 03/May/20

thank you sir, but what if I'm looking for log(i) in base 2 or 10?

Commented by mr W last updated on 03/May/20

$${\log }_{10}\left(i\right)=\frac{\ln \left(i\right)}{\ln 10}=\frac{\pi }{2\ln 10}i$$$${\log }_2\left(i\right)=\frac{\ln \left(i\right)}{\ln 2}=\frac{\pi }{2\ln 2}i$$$$...$$

Commented by MWSuSon last updated on 03/May/20

oh I see, change of base. Thank you sir.

Commented by mr W last updated on 03/May/20

$$i=e^{\frac{\pi i}{2}}$$$$\ln \left(i\right)=\ln \left(e^{\frac{\pi i}{2}}\right)=\frac{\pi }{2}i$$

Commented by MWSuSon last updated on 03/May/20

$$sirsothesolutionofthisequation$$$$2^{(x+1)}=i$$$$wouldproceedasfollows?$$$$x+1={log}_{2}i$$$$x={log}_{2}i-1$$$$x=\frac{lni}{ln2}-1$$$$x=\frac{\pi }{2ln2}i-1?$$$$$$

Commented by mr W last updated on 04/May/20

$$yes$$

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