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Solve Unknown Variable

  1. Jan 17, 2014 #1

    uml

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    1. The problem statement, all variables and given/known data
    The problem is attached.


    2. Relevant equations



    3. The attempt at a solution
    I've been trying to take the Log for both sides but the problem is I have (1+0.05X) if I just have e^-0.1 X, I will take Log for both sides and solve it but in this case I have (1+0.05X) too!!
    I have no idea how to solve this. This is not homework problem, my school will start next week, I am trying to prepare for math class by doing some problems.
    Thank you
     

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    Last edited: Jan 17, 2014
  2. jcsd
  3. Jan 17, 2014 #2

    Student100

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    The problem also appears unknown! :)

    Thats better.

    Try to take the natural log on both sides, is there a property of logs you can use? Show us how you went about your attempt so it's possible to see where you made an error.
     
    Last edited: Jan 17, 2014
  4. Jan 17, 2014 #3

    Mark44

    Staff: Mentor

    Here's the problem:
    ##2.5 X 10^{-6} = e^{-.1x}(1 + .05x)##

    Because of the x appearing in two places on the right side, taking logs (natural or otherwise) on both sides won't do any good. My guess is that you're supposed to come up with an approximate or graphical solution here, as opposed to an analytic solution.
     
  5. Jan 17, 2014 #4
    Yes. Mark44 is obviously correct. There is no analytic solution. They must be asking for an approximate solution. It looks like the exponential term on the rhs is going to dominate. If the equation were just 1.0x10-6=e-0.1x, what value of x would be required? This would be a good initial guess for the answer. You could then solve the equation by successive substitutions, starting with this initial guess. Can you think of a successive substitution scheme that would do the trick?

    Chet
     
  6. Jan 17, 2014 #5

    Ray Vickson

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    As Mark44 has stated, your problem is ##2.5 \times 10^{-6} = e^{-.1x}(1 + .05x)##. Typically, because the function involves ##x \times\exp(-kx)## you cannot hope for a simple solution, but must, instead, try for a numerical solution. You can draw a graph of both sides of the equation to see where the two curves cross, then home in on more accurate values of ##x## using numerous numerical methods, such as Newton's method.

    However, in this case a special function has been devised by mathematicians to handle such problems: the Lambert function. Maple gets two solutions:
    [tex] x_1 = - 20 - 10 W\left( -\frac{e^{-2}}{200000} \right) \doteq -19.99999323 \\
    x_2 = - 20 - 10 W\left(-1, -\frac{e^{-2}}{200000} \right) \doteq 150.4173817 [/tex]
    Here, ##W(z), \: W(k,z)## are two "branches" of the so-called Lambert W function, defined as solutions of the equation ##W \exp(W) = z##.

    Generally speaking, I would not recommend trying for a 'symbolic' solution like those two above, but would go directly to a numerical approach.
     
  7. Jan 17, 2014 #6

    Student100

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    Ahhh, I see now. I had assumed they wanted to solve for X even though it would leave the soultion in terms of X. The above suggestions look good to me.
     
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