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Please Help Natural Log Question!

  1. Oct 9, 2004 #1
    Please Help!! Natural Log Question!

    Here is the question that is bothering me:

    I need to solve this natural log for x. Please I need step by step instructions on how to figure out x. Thanks very much.

    ln (x) + ln (x+1) = 2
     
  2. jcsd
  3. Oct 9, 2004 #2

    Diane_

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    Remember the rules of logarithms: log(ab) = log a + log b

    That's kind of the original point of logs, actually, if you remember.

    So - do it backwards:

    ln x + ln(x + 1) = ln(x(x+1)) = ln(x^2 + x) = 2

    Does that get you far enough?
     
  4. Oct 9, 2004 #3

    Pyrrhus

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    Use logarithm properties

    [tex] \ln (AB) = \ln(A) + \ln(B) [/tex]
     
  5. Oct 9, 2004 #4
    Yes, I was able to get that far, but how do I get X all by itself on one side, like [ x = blah, blah]

    thanks
     
  6. Oct 9, 2004 #5

    Diane_

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    ln(a) = b implies a = e^b
     
  7. Oct 9, 2004 #6

    Pyrrhus

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    [tex] x^2 + x = e^2 [/tex]

    By e i mean antiln.
     
  8. Oct 9, 2004 #7
    yes, so it becomes e^2 = x^2 + x

    Now maybe my question is, what is the value of x, in order for that original equation to be true, x = ?

    Thanks
     
  9. Oct 9, 2004 #8

    Pyrrhus

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    Heh, Diane and myself seems to be helping almost in sync :rofl:
     
  10. Oct 9, 2004 #9

    Pyrrhus

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    Remember, Quadratic Equations.

    [tex] Ax^2 + Bx + C = 0[/tex]

    [tex] x^2 + x + (-e^2) = 0 [/tex]
     
  11. Oct 9, 2004 #10
    I finally got it, x = 2.26388......

    Thank you very much Cyclovenom and Diane, you've ended my day-long struggle (its pathetic i know, but its been a while) :grumpy:

    You guys are brilliant :blushing:
     
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