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This log problem is throwing me off, out of nowhere

  1. Nov 4, 2011 #1
    Question is: solve for x
    ln(x+1)-1 = ln(x-1)

    tried to do it TWO different ways, and neither is the right way, check this out:
    first attempt:
    since LN is on both sides , it goes away and you just have variables, that way sucks because then you get 0=-1 , wrong.
    second attempt:
    divide both sides by ln(x+1)
    that leaves -1 on one side and a bunch of nonsensical stuff on the other side.
    WHAT TO DO?
     
  2. jcsd
  3. Nov 4, 2011 #2

    Mentallic

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    Think about the formula ln(a)-ln(b) = ln(a/b) :wink:
     
  4. Nov 4, 2011 #3
    Your first assumption was not correct. Yet your subsequent reasoning was logical and you yourself noticed that the conclusion did not make sense.

    You do not have ln(....) = In(...)
    but you have In(...) + 1 = In(...).

    Follow the advice of Mentallic.
     
    Last edited: Nov 4, 2011
  5. Nov 4, 2011 #4
    Thats exactly what is above where the answer is wrong , where you get -1 on one side
     
  6. Nov 4, 2011 #5
    already did that and the answer was wrong, see above. answers not needed, but steps and how to
     
  7. Nov 4, 2011 #6
    -1=ln(x-1)-ln(x+1).
    Now try applying what Mentallic said. :wink:
     
  8. Nov 4, 2011 #7

    Mentallic

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    Is this what you're referring to?

    [tex]\ln\left(x+1\right)\neq \ln\left(x-1\right)[/tex]
    so you can't just take away the ln(...) from both sides to get 0=-1, use the formula I suggested and try again.
     
  9. Nov 4, 2011 #8

    SammyS

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    You can use each side of the equation,
    ln(x+1)-1 = ln(x-1)​
    as an exponent with a base of e.

    [itex]\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}[/itex]

    See what you can do with that.
     
  10. Nov 5, 2011 #9

    Mark44

    Staff: Mentor

    Alternatively, it might be simpler to write the first equation above as
    [itex]ln(x + 1) - ln(x - 1) = 1[/itex]

    and then use a property of logs to write the left side as ln of a single expression. Then exponentiate both sides.
     
  11. Nov 5, 2011 #10

    SammyS

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    Yes, Mark, I agree, but OP seemed to be unable to do this.
     
  12. Nov 5, 2011 #11

    Mentallic

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    But I"m failing to see how I could do anything with your suggestion other than to go back to where the question started. Can I get a hint? :smile:
     
  13. Nov 5, 2011 #12
    The simplest way to do it is to get it in the form
    In(x + 1) - In(x -1) = 1
    and then use In(a) - In(b) = In(a/b) as the other posters suggested.
     
  14. Nov 5, 2011 #13

    Mark44

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    There is no "In" function. It is ln (ell en), an abbreviation for natural logarithm (logarithm naturalis, in Latin).
     
  15. Nov 5, 2011 #14

    SammyS

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    Have I been asked what this suggestion was good for?

    [itex]\displaystyle e^{(\ln(x+1)-1)}=e^{\ln(x-1)}[/itex]

    is equivalent to:

    [itex]\displaystyle e^{\ln(x+1)}e^{-1}=e^{\ln(x-1)}[/itex]

    I hope OP can simplify [itex]\displaystyle e^{\ln(x+1)}\text{ and }e^{\ln(x-1)}[/itex]
     
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