# Ln(-1) = 0 ?!

by alba_ei
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 P: 38 supposed that we have $$\ln(-1)$$ then $$\frac{2}{2}\ln(-1)$$ so $$\frac{1}{2}\ln(-1)^2$$ this is equal to $$\frac{1}{2}\ln(1)$$ and if this is equal to 0 the we can say that $$ln(-1) = 0$$ is this right , wrong, are there any explanations for this?
 Mentor P: 8,316 The rule of logarithms is a lnx=lnxa. In this case, a=1-- you cannot split it into a fraction and then only take the numerator! If you look at the logarithm graph, you will see that the function is not defined for negative x.
 Sci Advisor HW Helper P: 2,020 It's wrong. ln(-1) is no longer a real number, so you can't treat it like one. This is like saying sqrt(-1) = (-1)1/2 = (-1)2/4 = ((-1)2)1/4 = 11/4 = 1.
HW Helper
P: 2,020
Ln(-1) = 0 ?!

 Quote by cristo The rule of logarithms is a lnx=lnxa. In this case, a=1-- you cannot split it into a fraction and then only take the numerator!
Actually, that step is perfectly valid in general - (a/a) ln(x) = 1/a ln(x^a), i.e. when everything is defined. Your next line explains why it's not valid here:
 If you look at the logarithm graph, you will see that the function is not defined for negative x.
Mentor
P: 8,316
 Quote by morphism Actually, that step is perfectly valid in general
Course it is; sorry!
P: 894
 Quote by cristo If you look at the logarithm graph, you will see that the function is not defined for negative x.
Really? Why can't one say that ln(-1)= i pi, for e^(i pi) = -1.
Or is there something wrong with that line of logic?
 P: 26 logarithm is defined also for complex numbers. ln(z)=ln(abs(z))+i*arg(z), where z is complex number, abs(z) is complex norm of complex number z, and arg(z) is its argument. So if -1 is treated as complex number -1+0*i, expression ln(-1) gives sense, but the identity a*ln(z)=ln(z^a) is no longer true.
 Mentor P: 8,316 To satisfy the pedants, I shall re-phrase my above answer. The natural logarithm function, whose argument is a real number and to whom we can apply the standard laws of logarithms, is not defined for negative real numbers.
 P: 617 for complex z: Ln(z) = ln(|z|) + i*Arg(z) so ln(-1) = ln(|-1|) + i*Arg(-1) = i*pi
HW Helper
P: 2,020
 Quote by JonF for complex z: Ln(z) = |z| + i*Arg(z) so ln(-1) = |-1| + i*Arg(-1) = 1 + i*pi
Really? If you were attempting to define the principal branch of Ln, then it ought to be Ln(z) = ln(|z|) + iArg(z), where ln is just the natural logarithm on the reals.

In this case, we have Ln(-1) = ln(|-1|) + iArg(z) = i*pi.
 P: 617 eh forgot the ln, fixed
Math
Emeritus
 HW Helper P: 3,348 $$\log_e -1 = i\pi + 2ki\pi, k\in \mathbb{Z}$$ Case Closed.
 Quote by Gib Z $$\log_e -1 = i\pi + 2ki\pi, k\in \mathbb{Z}$$ Case Closed.