Just like i^2 = -1 similarly, this problem needs ln(-1) x = ln(3) / ln(-3) = ln(3) / (ln(3) + ln(-1)) but -1 = e^(i pi) so ln(-1) = i pi Therefore, x = ln(3) / (ln(3) + i pi) and also ln(3) / (ln(3) + i pi + i 2 pi n) [ since e^(i 2 pi n) = 1, so ln(-1) = i pi + i 2 pi n for any integer n ]
Very cool
Just like i^2 = -1 similarly, this problem needs ln(-1)
x = ln(3) / ln(-3) = ln(3) / (ln(3) + ln(-1)) but -1 = e^(i pi) so ln(-1) = i pi
Therefore, x = ln(3) / (ln(3) + i pi) and also ln(3) / (ln(3) + i pi + i 2 pi n) [ since e^(i 2 pi n) = 1, so ln(-1) = i pi + i 2 pi n for any integer n ]