Results 1 to 3 of 3

Thread: problem with logarithms

  1. #1
    Member srirahulan's Avatar
    Joined
    Apr 2012
    From
    Srilanka
    Posts
    180

    Lightbulb problem with logarithms

    $\displaystyle log(x-2)+log2=2logy;\ log(x-3y+3)=0$ find out the value of (x) and (y).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: problem with logarithms

    Note that $\displaystyle \log(x-2) + \log 2 = \log(2x-4)$, and that $\displaystyle 2 \log y = \log y^2$. Therefore the first equation becomes

    $\displaystyle \log(2x-4) = \log y^2 \Rightarrow 2x-4 = y^2$

    The second equation, note that $\displaystyle \log 1 = 0$ (for any non-zero base), so $\displaystyle x - 3y + 3 = 0$.

    Solve the system of equations.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: problem with logarithms

    Whoops, $\displaystyle x-3y +3 = 1$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. word problem using logarithms
    Posted in the Business Math Forum
    Replies: 2
    Last Post: Dec 2nd 2009, 05:16 PM
  2. word problem using logarithms
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Dec 1st 2009, 11:01 AM
  3. logarithms problem
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Sep 2nd 2009, 11:16 PM
  4. Another problem in exponential and logarithms
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Jan 5th 2008, 07:48 AM
  5. Problem in exponential and logarithms
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Jan 5th 2008, 02:30 AM

Search Tags


/mathhelpforum @mathhelpforum