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!

Question about logarithms

  1. Jun 23, 2008 #1
    I have two problems if that's ok with you

    1. log((x-3)/(x+3)) <= 1

    2. 6*9^(1/x) - 13*6^(1/x) + 6*4^(1/x) = 0

    So for the first one i didnt get too far. I divided the logarithm and i got:
    log(x-3) + log(x+3) <= 1

    For the second one i used logarithms and received:
    log6 + 1/x*log9 - log13 - 1/x*log6 + log6 + 1/x*log4 = 0
    Next:
    1/x(log9-log6+log4) + log6 - log13 + log6 = 0
    Then:
    1/x*(log6) + log(36/13) = 0

    and i dont know what to do from here. any help would be apreciated. thanks :)
     
  2. jcsd
  3. Jun 23, 2008 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Let's start with the first one. Start by finding the value of y such that log(y)=1.

    You also need to be careful with your logarithm laws, your expansion is incorrect.
     
    Last edited: Jun 23, 2008
  4. Jun 23, 2008 #3
    Am I seeing that right? Is that log [(x-3)/(x+3)] <= 1

    And by <= you mean that it's less than one?

    Sorry guys. Getting used the forums here. I also thought that a good way to brush up my "skills" is to try and help other people.

    Thanks.
     
  5. Jun 23, 2008 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's how I read it.
    That generally means 'less than or equal to' (i.e. [itex]\leq[/itex]).
    Hey, no need to apologise. Welcome to the Forums! Be sure to introduce yourself in the General Discussion forum :wink:.
     
  6. Jun 23, 2008 #5
    Sorry it's just a typo. The sign in between is supposed to be a minus.

    Btw u helped me a lot with that hint so let me just chech if i got it.

    So we forget about the expansion and we have log[(x-3)/(x+3)] <= log10
    Next
    (x-3)/(x+3) <= 10
    x-3 <= 10x + 30

    x<= -33/9 << is this ok?
     
  7. Jun 23, 2008 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Are you sure that that should be a 10?
     
  8. Jun 23, 2008 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    As Hootenanny said, first solve log(y)= 1, then use y= (x-3)/(x+3). Remember than an inequality can change from "<" to ">" only at points where both sides are equal or where the function is not defined.

    \
    9= 32, 6= 3(2) and 4= 22. If you let a= 31/x and b= 21/x, this is 6a2- 13ab+ 6b2= 0. That's fairly easy to factor and so gives you two linear equations for a and b.

     
  9. Jun 23, 2008 #8
    Well if log(x) = 1
    then 10[tex]^{1}[/tex] = x

    From here x equals 10
    I think it is ok...
     
  10. Jun 23, 2008 #9
    I cant see an easy way to factor it, honestly. Is it just me?
     
  11. Jun 23, 2008 #10

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It is easy to factor, but if you don't see it you can always fall back on the quadratic formula. What are the solutions of [itex]6x^2-13x+6=0[/itex]?
     
  12. Jun 24, 2008 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Logarithms

    So (x-3)/(x+3)= 10. Can you solve that for x?
     
  13. Jun 24, 2008 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Logarithms

    Yes, it really is!. It might help you to know that 3(3)+ 2(2)= 13.
     
  14. Jun 24, 2008 #13

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Logarithms

    Sorry, my bad, I misread the OP.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question about logarithms
  1. Logarithmic question- (Replies: 5)

  2. Logarithms question (Replies: 5)

Loading...