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Log3(5x-4)+log3(2x+7) = 2

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    log3(5x-4)+log3(2x+7) = 2

    2. Relevant equations

    log3(5x-4)+log3(2x+7) = 2

    3. The attempt at a solution

    log3(5x-4)+log3(2x+7) = 2

    Find the value for x.....?
     
  2. jcsd
  3. Feb 2, 2009 #2

    Mark44

    Staff: Mentor

    Re: Log

    Your problem statement, relevant equation, and attempt all show the same thing. What have you actually tried to do?

    Look at the properties of logs, particularly the one that says log(a) + log(b) = log(ab).
     
  4. Feb 2, 2009 #3
    Re: Log

    When I did it that way, I get a equation with x squared.....can you have two values for x...??
     
  5. Feb 2, 2009 #4

    Mark44

    Staff: Mentor

    Re: Log

    Sure. Make sure though that any solutions of the quadratic are actually solutions of your log equation. For this problem, 5x - 4 must be positive. 2x + 7 also must be positive, but that will happen automatically if 5x - 4 > 0.
     
  6. Feb 2, 2009 #5
    Re: Log

    I did not quite get by ..any solutions of the quadratic are actually solutions of your log equation...
    Could you explain it in simple terms please
     
  7. Feb 2, 2009 #6

    Mark44

    Staff: Mentor

    Re: Log

    When you convert an expression such as log(a) + log(b) to log(ab), the assumption as that both a and b are positive. If you perform an operation that gets rid of the log function, it still must be that a and b are positive, even if that's not apparent in the equation you end up with.

    You have said that you ended up with a quadratic equation, which presumably you are going to solve for x. The solutions of your quadratic might or might not be solutions of the original log function.
     
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