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Homework Help: Unusual log problem

  1. Mar 24, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve for "x" in the equation: ln(x+2)-ln(x+3)+ln(x)=1

    3. The attempt at a solution

    This unusual ln problem really has me stumped with what to do and usually this stuff has given me no problems. I can't isolate the x because e is always in the way of solving it. From "ex+3e=x^2+2x" the answer I got was "-2+2(1+3e)^1/2/2e" (?)
  2. jcsd
  3. Mar 24, 2008 #2
    e is just a number, so ex+3e=x^2+2x is just a quadratic. put it in the form ax^2 + bx +c and apply the quadratic formula.
  4. Mar 24, 2008 #3
    I've already done that which simplifies to the answer shown in the first post. I set it equal to ex and then divided by e. I'm not clear as to why it equalled to zero at the end which can't possibly make any physical sense

    Another approach I used was to treat 2x-ex as "b" although that produced:


    That still makes it wrong (ugh)
    Last edited: Mar 24, 2008
  5. Mar 24, 2008 #4
    The answer in your first post is wrong. you'll have to give more details to tell what went wrong. You just want to collect terms with x^2, x and things indendent of x, and then to apply the quadratic formula.
  6. Mar 24, 2008 #5
    Take ln x to R.HS. and express 1 as 2.303*log10/2.303

  7. Mar 24, 2008 #6
    The quad formula doesn't seem to work in any of the cases and I'm extremely reluctant to change e to its actual irrational form because it'll just get messier.

    Physixguru, I'm scratching my head here - where did you get 1= 2.303*log10/2.303?

    Kamerling, for the first trial I thought about just isolating the x. So, ex=x^2+2x-3e. Then I tried x^2+2x-ex-3e where b is 2x-ex
    Last edited: Mar 24, 2008
  8. Mar 24, 2008 #7

    Gib Z

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    Take kammerlings original suggestion again. Treat e like any other constant, pretend you know nothing about its numerical value. You should get [itex]x^2 + (2-e) x - 3e = 0[/itex], a simple quadratic equation.
  9. Mar 24, 2008 #8
    ln x= 2.303 log x (assuming base 10 )

    now 1= 2.303/2.303 ( log 10)
  10. Mar 24, 2008 #9
    Here is how I did it this time- all the actual steps

    -2-e+( (2-e)^2-4(1)(-3e) )^1/2/2

    -2-e+( e^2-4e+4+12e )^1/2/2

    -2-e+( e^2+8e+4 )^1/2/2

    -2-e+( 4(1+2e) )^1/2/2



    Can someone confirm if this is right?
    Last edited: Mar 24, 2008
  11. Mar 24, 2008 #10


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    None of this makes any sense at all. Apparently "physixguru" is advocating changing to common logs. I have no idea why.
  12. Mar 24, 2008 #11


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    [-(2-e)+/- ( (2-e)^2-4(1)(-3e) )^1/2]/2= [-2+e+/- ( (2-e)^2-4(1)(-3e) )^1/2]/2
    [tex]\frac{-(2-e)\pm\sqrt{(2-e)^2- 4(1)(-3e)}}{2}= \frac{-2+e\pm\sqrt{e^2+ 8e+ 4}}{2}[/tex]
    Notice the sign difference.

    Last edited by a moderator: Mar 24, 2008
  13. Mar 24, 2008 #12
    Thanks HallsofIvy! :-) Made my day as usual
  14. Mar 24, 2008 #13
    would you like an accurate approach to this problem, ArtofScience?
  15. Mar 25, 2008 #14
    x = 3.2373049

    The answer is quite simple to obtain. You had it right when you stated xˆ2 + 2x = ex + 3e.

    Now just set the equation = 0, so xˆ2 + (2-e)x - 3e = 0...

    At this point, it's only a matter of graphing it on your calculator or plugging it into the quad form. Either way, you will still obtain the answer of 3.2373049. To check this answer, just plug it back into the original equation for x and be amazed at how it works!
  16. Mar 25, 2008 #15


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    In what sense is that more accurate than the solution ArtofScience gave?
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