1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: SImple Logarithms

  1. Jul 2, 2003 #1
    I need help on a lograrith HM:

    Solve for X: log(x-1)=log(x-2)-log(x+2)
    I got till here: log(x-1)=(log(x-2)/(x+2))

    Donot know what to do after wards

    ***Hey if you could tell me the answer please tell me how you derived it. i want to understand the problem. And Base if course to the 10.
    Last edited: Jul 2, 2003
  2. jcsd
  3. Jul 2, 2003 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hi anil,

    I'll give you a hint that should unlock the whole problem for you.


    You can apply that to the right side, and then take the antilog of both sides.
  4. Jul 2, 2003 #3
    Re: Re: SImple Logarithms

    :) ah tom, he has already done that
    its after that he is stuck :)

    to antilog:
    Since the bases are the same (log e), raise both sides to the power of e. This will eliminate the log e on both sides leaving you with a simple equation to solve.
  5. Jul 3, 2003 #4


    User Avatar
    Science Advisor

    To Tom: As Cummings pointed out, the anil had already done the
    log(x-1)= log((x-2)/(x+2)) part.

    To Cummings: anil also said "And Base if course to the 10."!

    Actually it doesn't matter: as long as a function is "one-to-one" (and logarithm to any base is one-to-one) if f(x)= f(y) then x= y.

    Since log(x-1)= log((x-2)/(x+2)) we have x-1= (x-2)/(x+2). Now multiply both sides by x+2 to get (x+2)(x-1)= x-2 so you have the quadratic equation x<sup>2</sup>+ x- 2= x-2. You ought to be able to solve that easily!
  6. Jul 3, 2003 #5

    Actually i solved it like this:

    log(x-1) = log(x-2)-log(x+2)
    log(x-1) = log((x-2)/log(x+2))
    10^log(x-1) = 10^log((x-2)/(x+2))
    (x-1) = (x-2)/(x+2)
    (x-1)(x+2) = (x-2)(x+2)/(x+2)
    x^2+2x-x-2 = (x-2)
    x^2+(x-2) = (x-2)
    x^2+(x-2)-(x-2) = (x-2)-(x-2)
    x^2 = 0
    sqrt(x^2) = sqrt(0)
    x = 0
    But if you Plug o back in Log(0-1) = not possible in real values? So thats is an extraneous solution?
  7. Jul 3, 2003 #6

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No fair, he edited that part in as I made my comment. :frown:
    (See how his edit and my post are 1 second apart?)
  8. Jul 3, 2003 #7
    Sorry Tom

    I posted it and then i was reading the rules. So it says that post it so that people know you tried. SO i didn't read the post you posted. SOrry tom =(
  9. Jul 6, 2003 #8


    User Avatar
    Science Advisor

    Okay, Tom, I see now.

    (You know it has never occurred to me to go back and edit my mistakes! I'll have to start doing that (and then pretend I didn't make a mistake from the start!)

    Anil, you are correct that the equation log(x-1)= log(x+2)- log(x+2) has NO real number solutions.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook