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Math Help - logs!

  1. #1
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    logs!

    Can anyone give me a hand with this one?

    loga(x) + loga(x - 2) = loga(x + 4)
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by rtwilton View Post
    Can anyone give me a hand with this one?

    loga(x) + loga(x - 2) = loga(x + 4)
    Hello rt,

    Remember \log_bm + \log_bn = \log_bmn

    \log_ax+\log_a(x-2)=\log_a(x+4)

    \log_ax(x-2)=\log_a(x+4)

    x(x-2)=x+4

    Can you finish from here?
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  3. #3
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    I think so. I distributed it out, got everything on one side and got:

    x^2 -3x - 4 = 0

    so x = -1 and x = 4

    but x cannot be a negative number right? so x=4 is the answer?
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  4. #4
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    Quote Originally Posted by rtwilton View Post
    I think so. I distributed it out, got everything on one side and got:

    x^2 -3x - 4 = 0

    so x = -1 and x = 4

    but x cannot be a negative number right? so x=4 is the answer?
    You can take the log of a negative number,
    but your answer is no longer "real", but complex.

    A logarithm is the inverse of a power.

    If y=10^x, then x=\log(y).

    10^x is always positive. As a result, you can only take the log of a positive number.

    However, 10 raised to an imaginary number can be negative, so the log of a negative number is imaginary.

    So long as you use restrict your domain to real numbers, the log of a negative number is undefined.
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