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Math Help - Solving equation involving e and ln

  1. #1
    Newbie hymnseeker's Avatar
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    Solving equation involving e and ln

    Solve lnx + ln(x-3) = 0

    So far my working is

    e^(lnx) + e^(ln(x-3)) = e^0

    x + (x-3) = 1

    and so on

    But this leads me to the incorrect solution. Where did I go wrong?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by hymnseeker View Post

    e^(lnx) + e^(ln(x-3)) = e^0

    Where did I go wrong?
    \ln(a) + \ln(b) = \ln(ab)
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  3. #3
    Senior Member vincisonfire's Avatar
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    You can't do what you did.
    What you can do is  a + b = c \implies e^{a+b} = e^c
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  4. #4
    Newbie hymnseeker's Avatar
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    How foolish of me to miss that.

    Thanks again.
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by hymnseeker View Post
    Solve lnx + ln(x-3) = 0

    So far my working is

    e^(lnx) + e^(ln(x-3)) = e^0

    x + (x-3) = 1

    and so on

    But this leads me to the incorrect solution. Where did I go wrong?
    Hello hymnseeker,

    Try this:

    \ln x + \ln (x-3)=0

    \ln x(x-3)=0

    x(x-3)=e^0

    x^2-3x=1
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