Results 1 to 5 of 5

Math Help - Logarithmic equation

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    65

    Logarithmic equation

    Greetings,
    The logarithmic equation below is given and it is required to find all values of the parameter a such that the equation has a single solution.

    What puzzles me is that I believe first of all we would need to ensure that these logarithms exist so I look for all values of a for which the arguments of the logarithms are positive and get that a ranges from (0;1).

    Now when I equal the discriminant to 0, I get a = 1, which should not be amonst the searched values of a, but is both because when I replace it in the equation I get truth and because one of the answers in the textbook is 1. Additionally, the other one states that every a < -1/3 works, which is in complete inconsistence with what I have for a.
    Attached Thumbnails Attached Thumbnails Logarithmic equation-lg.gif  
    Last edited by Logic; July 11th 2009 at 11:14 AM. Reason: Forgotten attachment
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,913
    Thanks
    1760
    Awards
    1
    First the argument of a logarithm function must be positive.
    a\in(0,1) insures that is true for both sides.

    Second, the logarithm function is injective (one-to-one) on its domain.
    Therefore solve 2x^2-4(a-2)+8+2a=x^2-2ax+a.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2008
    Posts
    65
    Yes, according to this a = 1 should not be a solution, but it is.
    Moreover, I think both arguments exist for a = 1.. ?
    And the answer given is:
    a = 1 or a < -1/3
    Neither of which fit in (0;1).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,913
    Thanks
    1760
    Awards
    1
    Quote Originally Posted by Logic View Post
    Yes, according to this a = 1 should not be a solution, but it is.
    No that is not correct, as I read the problem.
    If a=1 then x^2-2x+1=(x-1)^2.
    But (x-1)^2=0 if x=1.
    Therefore \log(x^2 -2ax+a) is not defined for x=1; i.e. for all x.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2008
    Posts
    65
    Well.
    I take the answer as incorrect in that case.

    Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Logarithmic equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 15th 2010, 09:14 AM
  2. Logarithmic Equation
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: May 10th 2010, 02:55 AM
  3. Logarithmic Equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 27th 2009, 02:41 AM
  4. Logarithmic equation
    Posted in the Algebra Forum
    Replies: 6
    Last Post: March 18th 2008, 10:17 AM
  5. Logarithmic Equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 12th 2007, 06:18 PM

Search Tags


/mathhelpforum @mathhelpforum