Can ln(-1) be equal to 0?

[alba_ei]

ln(-1) = 0 ?!

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?

 

[cristo]

The rule of logarithms is a lnx=lnx^a. 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.

 

[morphism]

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 = 1^1/4 = 1.

 

[morphism]

The rule of logarithms is a lnx=lnx^a. 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.

 

[cristo]

Actually, that step is perfectly valid in general

Course it is; sorry!

 

[JustinLevy]

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?

 

[Semo727]

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.

 

[cristo]

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.

 

[JonF]

for complex z: Ln(z) = ln(|z|) + i*Arg(z)

so ln(-1) = ln(|-1|) + i*Arg(-1) = i*pi

 

[morphism]

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.

 

[JonF]

eh forgot the ln, fixed

 

[HallsofIvy]

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?

Cristo said, "for negative x". Since the complex numbers are not an ordered field, there are no "negative" complex numbers. Cristo was clearly talking about real numbers.

 

[Gib Z]

$$\log_e -1 = i\pi + 2ki\pi, k\in \mathbb{Z}$$ Case Closed.

 

[murshid_islam]

$$\log_e -1 = i\pi + 2ki\pi, k\in \mathbb{Z}$$ Case Closed.

is the logarithm of complex numbers defined for any other base except 'e'?

 

[AlphaNumeric]

Yes, you can still change between various different bases for your logarithms in the same manner as you do for Real numbers.