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Question Number 58076 by Kunal12588 last updated on 17/Apr/19
 a^x =m  ⇒log_a m = x  So is following true  i^2 =−1  log_i (−1)=2
$$\:{a}^{{x}} ={m} \\ $$$$\Rightarrow\mathrm{log}_{{a}} {m}\:=\:{x} \\ $$$${So}\:{is}\:{following}\:{true} \\ $$$${i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\mathrm{log}_{{i}} \left(−\mathrm{1}\right)=\mathrm{2}\: \\ $$
Commented by $@ty@m last updated on 17/Apr/19
log_a m = x is defined only  for a(≠1)∈R
$$\mathrm{log}_{{a}} {m}\:=\:{x}\:{is}\:{defined}\:{only} \\ $$$${for}\:{a}\left(\neq\mathrm{1}\right)\in\mathbb{R}\: \\ $$
Answered by $@ty@m last updated on 17/Apr/19
Answered by MJS last updated on 17/Apr/19
log_i  (−1) =((ln (−1))/(ln i))  e^(iπ(2m+1)) =−1 ⇒ ln (−1) =iπ(2m+1)  e^(iπ(2n+(1/2))) =i ⇒ ln i =iπ(2n+(1/2))  ⇒ log_i  (−1) =((4m+2)/(4n+1)) with m, n ∈Z
$$\mathrm{log}_{\mathrm{i}} \:\left(−\mathrm{1}\right)\:=\frac{\mathrm{ln}\:\left(−\mathrm{1}\right)}{\mathrm{ln}\:\mathrm{i}} \\ $$$$\mathrm{e}^{\mathrm{i}\pi\left(\mathrm{2}{m}+\mathrm{1}\right)} =−\mathrm{1}\:\Rightarrow\:\mathrm{ln}\:\left(−\mathrm{1}\right)\:=\mathrm{i}\pi\left(\mathrm{2}{m}+\mathrm{1}\right) \\ $$$$\mathrm{e}^{\mathrm{i}\pi\left(\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)} =\mathrm{i}\:\Rightarrow\:\mathrm{ln}\:\mathrm{i}\:=\mathrm{i}\pi\left(\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\mathrm{log}_{\mathrm{i}} \:\left(−\mathrm{1}\right)\:=\frac{\mathrm{4}{m}+\mathrm{2}}{\mathrm{4}{n}+\mathrm{1}}\:\mathrm{with}\:{m},\:{n}\:\in\mathbb{Z} \\ $$

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