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Logarithmic equation

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data
    ##log _3\left(x-5\right) + log_3\left(x+3\right) = 2##

    I'm having trouble with applying properties of logarithms to solve equations, I think misunderstanding something fundamental here, I get most of the questions right but there are a handful that I have no idea what I am doing. Please help.
    2. Relevant equations

    3. The attempt at a solution
    I try to condense the logs into one with the product rule so:
    ##log_3\left(x^2-2x-15\right) = 2##

    then I rewrite it in exponential form:

    Now I figured that getting rid of the logs would make this quadratic equation manageable but I am coming up nowhere near the right answer (I peaked, its 6).

    If I add 15 to both sides I get:

    but I don't know what to do with the quadratic...I feel like I messed up trying to get rid of the logs but I cant really identify where I screwed up. I tried to rework it a couple times with different tricks but I still am nowhere close. Any help would be appreciated.
  2. jcsd
  3. Mar 18, 2014 #2


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    Quadratic formula?
  4. Mar 18, 2014 #3
    Well I didn't use the quadratic formula but I did try factoring it after subtracting 24 from both sides but I got negative roots and I thought you couldn't have those in a logarithms domain? Sorry... I know this is easy math for alot of you but I am teaching this to myself and am having a bit of trouble. Ill use the quadratic formula thanks.
  5. Mar 18, 2014 #4
    Just to check, can you please show us the factorisation you did after subtracting 24? There should be one positive and one negative root.
  6. Mar 18, 2014 #5


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    Why can't you use the quadratic formula?

    An alternative is simply to "complete the square":

    For example, take ##x^2 +2x - 3=0##. I want to write this as ##(x +1)^2 + F=0## for a certain ##F##. Clearly ##F=-4##. So the equation becomes ##(x+1)^2 -4 = 0##. Thus ##(x+1)^2 = 4##. Hence, ##x+1 = \pm 4##.
  7. Mar 18, 2014 #6
    Crap. I messed up on the factoring, I should have gotten (x-6)(x+4) but I wrote on my paper (x+6)(x+4). thanks.

    And I guess I just didn't think to use the quadratic formula(or completing the square), usually when I use it I end up with complex roots and I think for logs I have to have real non-zero solutions. I'm not sure, I just started learning logs yesterday.
  8. Mar 18, 2014 #7
    Oh and I have one more if you guys dont mind. This is one of the ones where I dont know where to start.
    1. The problem statement, all variables and given/known data
    ##log(x+4)= logx+log4##

    2. Relevant equations

    3. The attempt at a solution

    I used the product rule to condense the right side

    but after that I have no idea what to do...
  9. Mar 18, 2014 #8


    Staff: Mentor

    If the logs of two numbers are equal, then the two numbers are equal.

    Also, it's a good idea right at the start, to indicate what numbers are allowed for each of the log expressions. In this problem, you must have x > - 4 for log(x + 4) to be defined, and you must have x > 0 for log(x) to be defined. Together, these mean that x > 0.
  10. Mar 18, 2014 #9


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    Move the logs to one side :smile: Also keep in mind that [itex]a^0=1[/itex] for any positive a.
  11. Mar 18, 2014 #10
    Okay. So the equation becomes
    its not a quadratic so I cant factor it into two roots right? Im sorry I still dont know what to do from here...
  12. Mar 18, 2014 #11
    If I move them to one side I get

    ##log(4x/x+4)=0## because of the quotient property right? but I don't know how to simplify it further.
    And Im not sure I understand how I should use ##a^0=1##
    its seems like the terms have an implied power of 1.
  13. Mar 18, 2014 #12


    Staff: Mentor

    Right, it's not a quadratic. Add -x to both sides. This is really a simple equation...
  14. Mar 18, 2014 #13
    ##e^0 = ...##
  15. Mar 18, 2014 #14
    Alright sorry...
  16. Mar 18, 2014 #15


    Staff: Mentor

    You have really made it complicated. Also, I think the fraction is intended to be ##\frac{4x}{x + 4}##, so you should have written it as 4x/(x + 4) if that's what you meant. Without the parentheses, knowledgeable people would interpret it as (4x/x) + 4.
    Of course. x is the same as x1.
  17. Mar 19, 2014 #16


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    I woke up too early this morning to start answering questions. Clearly if

    [tex]\log(a)=\log(b)[/tex] then a=b. Moving everything to one side, using a log property and then using the definition of logs as so:






    Is just a long winded way of showing the same thing.
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