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a-x-m-log-a-m-x-So-is-following-true-i-2-1-log-i-1-2-




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
ax=mlogam=xSoisfollowingtruei2=1logi(1)=2
Commented by $@ty@m last updated on 17/Apr/19
log_a m = x is defined only  for a(≠1)∈R
logam=xisdefinedonlyfora(1)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
logi(1)=ln(1)lnieiπ(2m+1)=1ln(1)=iπ(2m+1)eiπ(2n+12)=ilni=iπ(2n+12)logi(1)=4m+24n+1withm,nZ

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