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Math Help - Natural Log Equation

  1. #1
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    Question Natural Log Equation

    I have the problem

    ln(c)+ln(2-c)=3
    I get c=2 or c=0. Then when I plug those values in, it's wrong. Is there even a solution? Is it undefined because of the -c?
    Thanks.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by dwatkins741 View Post
    I have the problem

    ln(c)+ln(2-c)=3
    I get c=2 or c=0. Then when I plug those values in, it's wrong. Is there even a solution? Is it undefined because of the -c?
    Thanks.
    Since logA+logB=log(AB) you can write

    log((2-c)c)=3

    c(2-c)=e^3

    finish it..
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  3. #3
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by dwatkins741 View Post
    I have the problem

    ln(c)+ln(2-c)=3
    I get c=2 or c=0. Then when I plug those values in, it's wrong. Is there even a solution? Is it undefined because of the -c?
    Thanks.
    \log(c)+\log(2-c)=3

    \implies \log{(c(2-c))}=3

    \implies c(2-c)=e^3

    \implies -c^2+ 2c- e^3 = 0

    Now, use the quadratic equation to find your solutions.

    \frac{-2 \pm \sqrt{2^2 - 4 (-1)(- e^3)}}{2(-1)}

    You may get a complex or two.
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  4. #4
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    Quote Originally Posted by dwatkins741 View Post
    I have the problem

    ln(c)+ln(2-c)=3
    I get c=2 or c=0. Then when I plug those values in, it's wrong. Is there even a solution? Is it undefined because of the -c?
    Thanks.
    First, there is no "-c". There is "2- c" which will be positive as long as c< 2. Second, c= 0 can't possibly be a solution specifically because ln is defined only for positive argument.

    Others have shown you how to solve that. Just out of curiosity how did you "get c= 2 or c= 0"? Did you set c= 0 and 2- c= 0?
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