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Math Help - Solving Log Equations

  1. #1
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    Solving Log Equations

    1.) xe^-2x + 2e^-2 = 0


    2.) x^2 ln x - 9 ln x = 0


    3.) 2 ln 3x - 1 = 1


    Thanks in advance for any help!
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  2. #2
    GAMMA Mathematics
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    The answer to the last one is \frac{e}{3}. Add the one to the right side; divide by two, and then exponentiate each side. Do you follow that?

    As far as the first two are concerned...

    2) (x^2-9)ln(x)=0

    Solve for x: x=1,3
    Last edited by colby2152; January 21st 2008 at 11:05 AM.
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  3. #3
    Junior Member Singular's Avatar
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    No3.

    <br /> <br />
\begin{gathered}<br />
  2\ln 3x - 1 = 1 \\ <br />
  \ln (3x)^2  - \ln (e) = \ln (e) \\ <br />
  \frac{{9x^2 }}<br />
{e} = e \\ <br />
  x = \frac{e}<br />
{3} \\ <br />
\end{gathered} <br /> <br />

    No2.

    <br />
\begin{gathered}<br />
  x^2 \ln x - 9\ln x = 0 \\ <br />
  (x^2  - 9)\ln x = \ln 1 \\ <br />
  \ln x^{(x^2  - 9)}  = \ln 1 \\ <br />
  x^{(x^2  - 9)}  = 1 \\ <br />
  x = 1,3\\ <br />
\end{gathered} <br /> <br />
    Last edited by Singular; January 23rd 2008 at 08:38 PM.
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by colby2152 View Post
    2) (x^2-9)ln(x)=0

    Solve for x: x=-3,0,3
    Little mistake here

    Central solution should be x=1.

    Quote Originally Posted by toop View Post
    3.) 2 ln 3x - 1 = 1
    This is \ln3x=1. Can you take it from there?

    Anyway use parenthesis please, it could be 2\ln(3x-1)=1. (See my signature.)
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  5. #5
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    Quote Originally Posted by colby2152 View Post
    The answer to the last one is \frac{e}{3}. Add the one to the right side; divide by two, and then exponentiate each side. Do you follow that?

    As far as the first two are concerned...

    2) (x^2-9)ln(x)=0

    Solve for x: x=-3,0,3
    Hi,

    only a short notice: The ln()-function is not definde for negative numbers. Therefore x = -3 is not a solution of this equation.
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  6. #6
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    Oops! I posted the problem wrong for #3... It should have been

    3.) 2 ln 3x = 1

    Sorry about that!
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  7. #7
    GAMMA Mathematics
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    I was even thinking that when I did the problem, but I still kept x = 0 as a solution. To the OP, you cannot take the natural log of a number zero or less.
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