Results 1 to 12 of 12

Thread: Logarithms; negative numbers and its answers?

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    6

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    680
    Thanks
    19
    Quote Originally Posted by MAtsOvechkin View Post
    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.
    Last edited by adkinsjr; Nov 8th 2009 at 12:22 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by MAtsOvechkin View Post
    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 $\displaystyle \mathbb{C}$

    $\displaystyle log(-x) = log(x) + i\pi$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2009
    Posts
    6
    Quote Originally Posted by adkinsjr View Post
    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.

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

    $\displaystyle log(-x) = log(x) + i\pi$
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by MAtsOvechkin View Post
    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

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

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

    From Euler's identity $\displaystyle ln(-1) = i \pi$ so $\displaystyle log(-1) = \frac{i\pi}{ln(10)} $ which is where the 1.36... comes from
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2009
    Posts
    6
    Quote Originally Posted by e^(i*pi) View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by MAtsOvechkin View Post
    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 $\displaystyle \mathbb{C}$ seems to use $\displaystyle e$ a lot

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

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

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

    When b=10 you get $\displaystyle \frac{i\pi}{ln(10)}$ and $\displaystyle \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 $\displaystyle log_{10}(-6)$

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

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

    Remembering to divide by ln(10) gives

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

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

    EDIT II: The basic formula is $\displaystyle ln(-x) = ln(x) + i\pi$. I would suggest using the change of base rule to put into the form $\displaystyle a\,ln(-x)$ first where $\displaystyle a$ is a constant
    Last edited by e^(i*pi); Nov 8th 2009 at 12:51 PM. Reason: See post
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Nov 2009
    Posts
    6
    Quote Originally Posted by e^(i*pi) View Post
    The basic definition followed by the change of base rule since $\displaystyle \mathbb{C}$ seems to use $\displaystyle e$ a lot

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

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

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

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

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

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

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

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

    Remembering to divide by ln(10) gives

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

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

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

  9. #9
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by MAtsOvechkin View Post
    Thank you, you have completely answered my question!
    Thank you for expanding my knowledge on complex logarithms that are not base e
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Nov 2009
    Posts
    6
    Quote Originally Posted by e^(i*pi) View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    I used google as my calculator and the inspiratation came from How to find the log of a negative number? - Yahoo! UK & Ireland Answers
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Nov 2009
    Posts
    6
    Quote Originally Posted by e^(i*pi) View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Feb 26th 2010, 05:27 PM
  2. negative fractional exponents and logarithms
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Feb 24th 2010, 05:46 AM
  3. negative numbers
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Oct 19th 2008, 04:05 AM
  4. negative numbers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jul 28th 2008, 04:10 AM
  5. Negative numbers squared
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Jun 1st 2008, 04:05 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum