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The value for t

  1. Oct 15, 2016 #1
    • Member warned that an effort must be shown
    1. The problem statement, all variables and given/known data

    s = 0.5(e^t - t - 1)
    t=?
    when s=18
    2. Relevant equations
    the use of natural log probably

    3. The attempt at a solution
    I tried using the natural log rule but im getting stuck. Please help me out here. If someone could give a step by step solution,I'll be rreally glad, Cheers
     
  2. jcsd
  3. Oct 15, 2016 #2

    Math_QED

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    It's a hard forum requirement that you show some work, what you tried, what you thought, etc. Nobody here will write out the entire solution for you.
     
  4. Oct 15, 2016 #3
    @Math_QED

    18=0.5e^t - t -1

    19 = 0.5e^t -t

    ln(19) = tln(0.5)-lnt ??
     
  5. Oct 15, 2016 #4

    Krylov

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    Are you looking for only positive solutions for ##t##? How many solutions do you think there are in total? Make a graph.
    It is probably not possible to find an expression for ##t## in terms of standard functions, so you may have to call upon a numerical root finder.
     
  6. Oct 15, 2016 #5
    Here, t represents time so if theres a easier way to find only the positive values, then i'd like to know that please :(
     
  7. Oct 15, 2016 #6

    Krylov

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    There is no solution (positive or otherwise) in terms of elementary functions, so if this is a homework problem that asks you to find an exact solution, you may have to re-read the problem.
    If this is indeed the correct equality, then you could proceed to make a graph to get an idea of where the solution may be. Then you can invoke a root finding to determine it approximately.
     
  8. Oct 15, 2016 #7
    upload_2016-10-15_16-56-12.png @Krylov this is the solution. but im not sure how he got to the value of t :/
     
  9. Oct 15, 2016 #8
    Alright sorry i didnt put the bracket. My apologies :p
     
  10. Oct 15, 2016 #9

    Krylov

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    No, ##t = 3.7064## does not satisfy ##18 = 0.5(e^t - t - 1)##. (Try it.) It satisfies this equality only approximately and it was probably found using a numerical root finder or by reading it off from a sufficiently detailed graph.
     
  11. Oct 15, 2016 #10
    What the heck?! I've wasted pages just to get the answer but i couldnt. I'll sue the author for this! lol . I'll take this question to my dynamics teacher and see what he comes up with. Thanks man! @Krylov
     
  12. Oct 15, 2016 #11

    pasmith

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    You can't solve [tex]
    e^t - t = 37[/tex] analytically; if you have a calculator, then the iteration [tex]
    t_{n+1} = \ln(t_n + 37)[/tex] with [itex]t_1 = 0[/itex] converges rapidly to [itex]t = 3.706384959[/itex] to the precision my calculator will display. (How did I know to try that instead of [itex]
    t_{n+1} = e^{t_n} - 37[/itex]? For the iteration [itex]t_{n+1} = f(t_n)[/itex] to converge to a solution of [itex]t = f(t)[/itex] which you know to exist then you must have [itex]|f'(t)| < 1[/itex] near that solution. [itex]\ln(t + 37)[/itex] has derivative [itex]0 < 1/(t + 37) < 1[/itex] for all [itex]t > 0[/itex]; [itex]e^t - 37[/itex] has derivative [itex]e^t > 1[/itex] for all [itex]t > 0[/itex].)
     
  13. Oct 15, 2016 #12

    Ray Vickson

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    No, absolutely NOT.
    $$\ln(0.5 e^t - t) \neq t \ln(0.5) - \ln(t)$$.
    Never, ever write $$\ln(a \pm b) = \ln(a) \pm \ln(b) \; \Longleftarrow \; \text{FALSE!}$$
    The correct property of logarithms is
    $$ \ln(a \times b) = \ln(a) + \ln(b)\; \Longleftarrow \; \text{True!}$$
    Anyway, this problem has no solution in elementary functions, but it can be solved in terms of the so-called Lambert W-function. That is a decidedly non-elementary function, whose evaluation will not appear on any calculator button and probably not even in any spreadsheet.

    It is probably easiest to just look for a numerical solution using one of the many numerical root-finding methods available.
     
  14. Oct 15, 2016 #13
    But i just used the natural log power rule
    ln(x^y) = y ∙ ln(x)
    how can that be wrong?
    @Ray Vickson
     
  15. Oct 15, 2016 #14

    Ray Vickson

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    No, you did not; I was responding to your post where you wrote

    19 = 0.5e^t -t

    ln(19) = tln(0.5)-lnt ??


    That was a "cut and paste" of your actual writing.

    Basically, you were saying that ##\ln(.5 e^t -t) = \ln(0.5 e^t) - \ln(t)##, which is definitely wrong.
     
  16. Oct 15, 2016 #15

    Mark44

    Staff: Mentor

    Noted.
     
  17. Oct 16, 2016 #16
    @Ray Vickson
    Oh i get it! Yeah you're right. Natural log really confuses me even though im well aware of the existence of these simple rules :p
     
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