1. ## Logarithms; negative numbers and its answers?

How do you take a log of a negative number? For example, what is the answer for Log -1? Log -0.08? Log -1.5?

Can you please show me the steps that need to be taken in order to take a log of a negative number, and what mathematical theory (IE: base formula of a log, etc) you used in order to get the value of a negative logarithm? Thanks

2. Originally Posted by MAtsOvechkin
How do you take a log of a negative number? For example, what is the answer for Log -1? Log -0.08? Log -1.5?

Can you please show me the steps that need to be taken in order to take a log of a negative number, and what mathematical theory (IE: base formula of a log, etc) you used in order to get the value of a negative logarithm? Thanks
The argument of a logarithmic function can never be negative.

EDIT: The argument can never be negative in the real numbers.

3. Originally Posted by MAtsOvechkin
How do you take a log of a negative number? For example, what is the answer for Log -1? Log -0.08? Log -1.5?

Can you please show me the steps that need to be taken in order to take a log of a negative number, and what mathematical theory (IE: base formula of a log, etc) you used in order to get the value of a negative logarithm? Thanks
There are no real solutions but there are in $\mathbb{C}$

$log(-x) = log(x) + i\pi$

The argument of a logarithmic function can never be negative.
It can be, but the solution would be in Complex. That's what I am wondering.

Originally Posted by e^(i*pi)
There are no real solutions but there are in $\mathbb{C}$

$log(-x) = log(x) + i\pi$
Thanks. So, if I follow this form, let's say I am calculating log(-4):
Therefore,
log(-4) = log4 + i/pi
log(-4) (approximately)= 0.602 + i/pi
Is this the way to calculate it? What would the value of i/pi be here? My calculator states that it would be approximately 1.36i, but I don't understand how it managed to multiply (sqrt) -1 by 3.14

5. Originally Posted by MAtsOvechkin
It can be, but the solution would be in Complex. That's what I am wondering.

Thanks. So, if I follow this form, let's say I am calculating log(-4):
Therefore,
log(-4) = log4 + i/pi
log(-4) (approximately)= 0.602 + i/pi
Is this the way to calculate it? What would the value of i/pi be here? My calculator states that it would be approximately 1.36i, but I don't understand how it managed to multiply (sqrt) -1 by 3.14
I will have to check that. Wikipedia says something about using Arg(z) but I don't follow it in your example

Edit: I believe it's derived from the change of base rule

$
log_{10}(-x) = log_{10}(x) + log(-1)$

$log(-1) = \frac{ln(-1)}{ln(10)}$

From Euler's identity $ln(-1) = i \pi$ so $log(-1) = \frac{i\pi}{ln(10)}$ which is where the 1.36... comes from

6. Originally Posted by e^(i*pi)
I will have to check that. Wikipedia says something about using Arg(z) but I don't follow it in your example
I am just wondering in general. Whenever I take a log of a negative number, the answer is always in the form of:

log (-x) = log(x) + 1.364376i.

How is the latter term (1.364376i) obtained? I have done multiplying i by pi, but that just gives me 3.14i. Do you know (or anyone else know) a site where I can see the proof of a logarithm with a negative arguement? Thanks again.

7. Originally Posted by MAtsOvechkin
I am just wondering in general. Whenever I take a log of a negative number, the answer is always in the form of:

log (-x) = log(x) + 1.364376i.

How is the latter term (1.364376i) obtained? I have done multiplying i by pi, but that just gives me 3.14i. Do you know (or anyone else know) a site where I can see the proof of a logarithm with a negative arguement? Thanks again.
The basic definition followed by the change of base rule since $\mathbb{C}$ seems to use $e$ a lot

$log_b(-x) = log_b(x) + log_b(-1)$

$log_b(-1) = \frac{ln(-1)}{ln(b)}$

$ln(-1) = i\pi$ which is where the pi part comes from

When b=10 you get $\frac{i\pi}{ln(10)}$ and $\frac{\pi}{ln(10)} = 1.364376
$

^ EDIT III: changed log(10) to ln(10) as it should be ln
-----------------------------

EDIT: I'll try to test by solving $log_{10}(-6)$

$log_{10}(-6) = \frac{ln(-6)}{ln(10)}$

As $ln(-x) = ln(x) + i\pi$ then we get $ln(-6) = ln(6) + i\pi$

Remembering to divide by ln(10) gives

$\frac{ln(6) + i\pi}{ln(10)} = \frac{ln(6)}{ln(10} + \frac{\pi}{ln(10)}\,i = log_{10}(6) + \frac{\pi}{ln(10)}\,i$

-----------------------------

EDIT II: The basic formula is $ln(-x) = ln(x) + i\pi$. I would suggest using the change of base rule to put into the form $a\,ln(-x)$ first where $a$ is a constant

8. Originally Posted by e^(i*pi)
The basic definition followed by the change of base rule since $\mathbb{C}$ seems to use $e$ a lot

$log_b(-x) = log_b(x) + log_b(-1)$

$log_b(-1) = \frac{ln(-1)}{ln(b)}$

$ln(-1) = i\pi$ which is where the pi part comes from

When b=10 you get $\frac{i\pi}{log(10)}$ and $\frac{\pi}{log(10)} = 1.364376
$

-----------------------------

EDIT: I'll try to test by solving $log_{10}(-6)$

$log_{10}(-6) = \frac{ln(-6)}{ln(10)}$

As $ln(-x) = ln(x) + i\pi$ then we get $ln(-6) = ln(6) + i\pi$

Remembering to divide by ln(10) gives

$\frac{ln(6) + i\pi}{ln(10)} = \frac{ln(6)}{ln(10} + \frac{\pi}{ln(10)}\,i = log_{10}(6) + \frac{\pi}{ln(10)}\,i$

-----------------------------

EDIT II: The basic formula is $ln(-x) = ln(x) + i\pi$. I would suggest using the change of base rule to put into the form $a\,ln(-x)$ first where $a$ is a constant
Thank you, you have completely answered my question!

9. Originally Posted by MAtsOvechkin
Thank you, you have completely answered my question!
Thank you for expanding my knowledge on complex logarithms that are not base e

10. Originally Posted by e^(i*pi)
Thank you for expanding my knowledge on complex logarithms that are not base e
By the way, did you have a certain website as a reference? If so, could you provide a link?

11. I used google as my calculator and the inspiratation came from How to find the log of a negative number? - Yahoo! UK & Ireland Answers

12. Originally Posted by e^(i*pi)
I used google as my calculator and the inspiratation came from How to find the log of a negative number? - Yahoo! UK & Ireland Answers
Thanks again