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Math Help - Logarithms; negative numbers and its answers?

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    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
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    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; November 8th 2009 at 12:22 PM.
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    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 \mathbb{C}

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

    <br />
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
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    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.
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    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 \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<br />

    ^ 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
    Last edited by e^(i*pi); November 8th 2009 at 12:51 PM. Reason: See post
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    Quote Originally Posted by e^(i*pi) View Post
    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<br />

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

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