Results 1 to 2 of 2

Math Help - logarithm problem

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    202

    logarithm problem

    how to solve this question


    <br /> <br />
2+\log_2{(3x-1)}=\log_2{4x}<br />
    Last edited by mastermin346; March 1st 2010 at 06:58 AM. Reason: wrong write
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by mastermin346 View Post
    how to solve this question


    <br /> <br />
2+\log_2{(3x-1)}=\log_2{4x}<br />
    1. Re-write the equation:

    2+\log_2{(3x-1)}=\log_2{4x}~\implies~2=\log_2{4x} - \log_2{(3x-1)}~\implies~ 2=\log_2\left( \frac{4x}{(3x-1)}\right)

    2. Now use the base 2 on both sides of the equation:

    4=\frac{4x}{(3x-1)} Solve for x. Keep in mind that x > \tfrac13.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Logarithm problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 5th 2010, 12:39 AM
  2. logarithm problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 8th 2009, 07:57 PM
  3. help with this logarithm problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 6th 2008, 06:07 PM
  4. logarithm problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: December 20th 2007, 05:54 AM
  5. Please help logarithm problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 1st 2006, 10:32 PM

Search Tags


/mathhelpforum @mathhelpforum