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Homework Help: Logarithm =/

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    solve
    ln(x+2)-ln(x+1)=1


    2. Relevant equations
    log laws


    3. The attempt at a solution
    hi there, well i tried to solve it but got stuck pretty much at the start
    => ln(x+2)/ln(x+1)=1
    =>ln(x+2)/ln(x+1)= e^1
    multiply out brackets and rearrange them?
    is this what i should do next
    OH wait does it become =>(x+2)/(x+1)=e , as ln disappears due to relation of y=lnx : x=e^y?
    Thanks in Advance.
     
    Last edited: Dec 8, 2009
  2. jcsd
  3. Dec 8, 2009 #2
    well, Ln A - Ln B = Ln (A/B)

    Then if Ln (A/B) = 1, you can anti log both sides, and solve for x

    so A/B = e^1

    for example
     
  4. Dec 8, 2009 #3
    oh yea, thanks alot i got it.. but ugh i got stuck on another equation this time =/
    its ln(x+3)+ln(x-1)= 0
    my attempt:
    since its log a+log b= log ab

    ln(x^2+2x-3)= o
    x^+2x-3= e^0 which is 1.
    is this correct?
     
  5. Dec 8, 2009 #4
    Yes but you have not solved for x yet.
     
  6. Dec 8, 2009 #5

    Mark44

    Staff: Mentor

    So far, so good, but you're not done.
    ln(x^2 + 2x - 3) = 0
    x^2 + 2x -3 = 1
    x^2 + 2x -4 = 0
    Now solve the quadratic. Keep in mind that for your original log expressions to be defined, x > - 3 and x > 1, which means that x > 1. If you get a value of x such that x <= 1, you have to discard it.
     
  7. Dec 8, 2009 #6
    thanks for your reply, after factorization the values i get are x=0,-4 .
    i don't quite get it, now what to do with the original expression =?
     
  8. Dec 8, 2009 #7

    Mark44

    Staff: Mentor

    Your factorization is incorrect.
     
  9. Dec 8, 2009 #8
    *facepalm* oh no =/ lol , sorry give me a sec
    x= -2+[tex]\sqrt{}5[/tex],-2-[tex]\sqrt{}5[/tex]
     
    Last edited: Dec 8, 2009
  10. Dec 8, 2009 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Still not quite right. Can you check that once more? And you should check the roots in your original equation. One or both of them may not be valid solutions.
     
  11. Dec 8, 2009 #10
    ah this is embarrassing its basic factorization =/ ,oo i dont know what i am doing wrong , i tried both methods to factorize it, sorry its just that my mind is not with me --> 3.52 am.
     
  12. Dec 8, 2009 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Use the quadratic formula. I assume you were doing that. You just got a number wrong. And again, don't forget to try and plug the roots back into the original equation and check that they actually work. You can get false roots.
     
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