1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    ##3^2=x^2-2x-15##

    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:
    ##24=x^2-2x##

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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
    ##log(x+4)=log(4x)##

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

    Mark44

    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

    Mentallic

    User Avatar
    Homework Helper

    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
    ##x+4=4x##
    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

    Mark44

    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

    Mark44

    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

    Mentallic

    User Avatar
    Homework Helper

    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:

    [tex]\log(a)=\log(b)[/tex]

    [tex]\log(a)-\log(b)=0[/tex]

    [tex]\log\left(\frac{a}{b}\right)=0[/tex]

    [tex]\frac{a}{b}=1[/tex]

    [tex]a=b[/tex]

    Is just a long winded way of showing the same thing.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Logarithmic equation
  1. Logarithmic equation (Replies: 10)

  2. Logarithm equation (Replies: 17)

  3. Logarithm equation (Replies: 8)

  4. Logarithmic Equation (Replies: 9)

  5. Logarithmic equation (Replies: 3)

Loading...