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 ]
Love your use of Euler's identity. A beautiful and simple argument which provides all the complex solutions to the problem, and moreover, also happens to be correct. Good job!
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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 ]
Love your use of Euler's identity. A beautiful and simple argument which provides all the complex solutions to the problem, and moreover, also happens to be correct. Good job!
Much simpler !
Very cool
(-3)^x
If we keep x A= 2/2
Where (-3)^2 = 9
Then root 9 is 3
So x is 1