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Solving a logarithmic equation.

  1. Oct 26, 2012 #1
    1. The problem statement, all variables and given/known data
    log3x+log3(2x+1)=1
    Solve for x.


    2. Relevant equations



    3. The attempt at a solution
    log3((2x+1)x)=1
    3log3((2x+1))=31
    (2x+1)x=3
    2x2+x=3
    2x2+x-3=0

    Using the quadratic formula
    x= (-1±√(12-4(2)(-3)))/(2(2))
    = (-1±√(1+24))/(4)
    = (-1±√25)/(4)
    = (-1±5)/(4)
    = 4/4 or -6/4

    Is there no solution?
     
  2. jcsd
  3. Oct 26, 2012 #2

    SteamKing

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    What's wrong with the two solutions you got from solving the quadratic eqn.?
     
  4. Oct 26, 2012 #3
    Logarithms with negative values are undefined though aren't they? So only 4/4 or 1 would be correct?
     
  5. Oct 26, 2012 #4

    Mark44

    Staff: Mentor

    You should simplify answers such as these. You have x = 1 or x = -3/2. There is still some work to do, though.
     
  6. Oct 26, 2012 #5

    SammyS

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    Yes, the other answer is what's called an extraneous solution.
     
  7. Oct 26, 2012 #6
    If I have x =1 and x =-3/2 and I know that -3/2 is an extraneous answer, then x = 1. I am not certain what else there is to do.
     
  8. Oct 27, 2012 #7

    Mark44

    Staff: Mentor

    If you've checked your solution and it works, you're done.
     
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