Results 1 to 7 of 7

Thread: Solving Log Equations

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    36

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Thanks
    1
    Awards
    1
    The answer to the last one is $\displaystyle \frac{e}{3}$. 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) $\displaystyle (x^2-9)ln(x)=0$

    Solve for x: $\displaystyle x=1,3$
    Last edited by colby2152; Jan 21st 2008 at 11:05 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Singular's Avatar
    Joined
    Dec 2006
    From
    Solo, Java
    Posts
    57
    No3.

    $\displaystyle

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

    $

    No2.

    $\displaystyle
    \begin{gathered}
    x^2 \ln x - 9\ln x = 0 \\
    (x^2 - 9)\ln x = \ln 1 \\
    \ln x^{(x^2 - 9)} = \ln 1 \\
    x^{(x^2 - 9)} = 1 \\
    x = 1,3\\
    \end{gathered}

    $
    Last edited by Singular; Jan 23rd 2008 at 08:38 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Quote Originally Posted by colby2152 View Post
    2) $\displaystyle (x^2-9)ln(x)=0$

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

    Central solution should be $\displaystyle x=1.$

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

    Anyway use parenthesis please, it could be $\displaystyle 2\ln(3x-1)=1.$ (See my signature.)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    Quote Originally Posted by colby2152 View Post
    The answer to the last one is $\displaystyle \frac{e}{3}$. 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) $\displaystyle (x^2-9)ln(x)=0$

    Solve for x: $\displaystyle x=-3,0,3$
    Hi,

    only a short notice: The ln()-function is not definde for negative numbers. Therefore x = -3 is not a solution of this equation.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2007
    Posts
    36
    Oops! I posted the problem wrong for #3... It should have been

    3.) 2 ln 3x = 1

    Sorry about that!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Thanks
    1
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Nov 30th 2011, 01:41 AM
  2. Solving two equations
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Nov 21st 2011, 02:55 PM
  3. Solving equations
    Posted in the Algebra Forum
    Replies: 29
    Last Post: Nov 18th 2011, 06:05 AM
  4. Solving these equations...
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Apr 27th 2009, 05:06 AM
  5. solving equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 16th 2006, 03:58 PM

Search Tags


/mathhelpforum @mathhelpforum