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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

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    Hi anil,

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

    log(a)-log(b)=log(a/b)

    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
    log(x-1)=(log(x-2)/(x+2))
    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

    HallsofIvy

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    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
    Yeah

    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

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    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

    HallsofIvy

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    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.
     
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