Results 1 to 2 of 2

Math Help - Logarithm

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    Spore
    Posts
    21

    Logarithm

    What's wrong with the workings in the attachment? Why can't I factorise?
    Attached Thumbnails Attached Thumbnails Logarithm-image.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32

    Re: Logarithm

    You're lucky I can read upside down.
    I would advise in the future that you become acquainted with the use of latex, even if it's only a rudimentary understanding like mine to avoid attachments and to make the problem legible. We have

    \log_3(4x-6) - \log_3(2) = \log_3(x^2-1)
    \log_3(2x-3) = \log_3(x^2-1)
    2x-3 = x^2-1
    x^2-2x+2=0

    You have done nothing wrong, if I have read what you've written correctly. Here we can use the quadratic formula to find the solutions for x directly, no need to factorize.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 18th 2011, 07:04 AM
  2. Logarithm
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 4th 2010, 04:53 PM
  3. logarithm help
    Posted in the Algebra Forum
    Replies: 7
    Last Post: January 2nd 2010, 08:26 AM
  4. Logarithm help
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: January 14th 2008, 02:54 PM
  5. logarithm
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 1st 2008, 08:24 AM

Search Tags


/mathhelpforum @mathhelpforum