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Question Number 58076 by Kunal12588 last updated on 17/Apr/19

$$a^x=m$$$$\Rightarrow {\log }_{a}m=x$$$$Soisfollowingtrue$$$$i^2=-1$$$${\log }_i(-1)=2$$

Commented by $@ty@m last updated on 17/Apr/19

$${\log }_{a}m=xisdefinedonly$$$$fora(\ne 1)\in \mathbb{R}$$

Answered by $@ty@m last updated on 17/Apr/19

Answered by MJS last updated on 17/Apr/19

$${\log }_{\mathrm{i}}(-1)=\frac{\ln (-1)}{\ln \mathrm{i}}$$$${\mathrm{e}}^{\mathrm{i}\pi (2m+1)}=-1\Rightarrow \ln (-1)=\mathrm{i}\pi (2m+1)$$$${\mathrm{e}}^{\mathrm{i}\pi (2n+\frac{1}{2})}=\mathrm{i}\Rightarrow \ln \mathrm{i}=\mathrm{i}\pi (2n+\frac{1}{2})$$$$\Rightarrow {\log }_{\mathrm{i}}(-1)=\frac{4m+2}{4n+1}\mathrm{with}m,n\in \mathbb{Z}$$

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