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Solve the following equatiom

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data

    In the formula theta = Ve^(-Rt/L), the value of theta = 58, V = 255, R = 0.1 and L = 0.5. Find the corresponding value of t.

    2. Relevant equations




    3. The attempt at a solution
    Rt/L = V - θ



    t = (L/R)(V - θ)



    t = (0.5/0.1)(255 - 58) = 985



    t = 985

    anywhere near?
     
  2. jcsd
  3. May 14, 2013 #2

    DrClaude

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    Staff: Mentor

    Where did that come from???
     
  4. May 14, 2013 #3
    think I did that incorrectly should it be:
    θ = Ve-Rt/ L

    Where, θ = 58, V = 255, R = 0.1, and L = 0.5.

    58 = 255 e-.01 t/0.5

    e-t/5 = 4.3966 (approx)

    Taking log on both sides

    -t/5 = ln(4.3966)

    -t = 7.40415

    t = -7.40415
     
  5. May 14, 2013 #4

    DrClaude

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    Staff: Mentor

    That line is not correct.
     
  6. May 14, 2013 #5
    I think at that point I should divide by 255

    so

    e^0.1t/0.5=0.2274

    then turn in to logarithmic statement

    0.1t/0.5=In0.2274

    then
    0.1t=0.5In0.2274

    then
    t=0.5In0.2274/0.1

    right track?
     
  7. May 14, 2013 #6

    DrClaude

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    Staff: Mentor

    Be careful with the minus signs. And pay attention to your notation, as that last line looks like
    $$
    t = 0.5 \ln \frac{0.2274}{0.1}
    $$
     
  8. May 14, 2013 #7
    as a point how would I correctly write that last part to avoid confusion?

    and after that I get

    t=0.5In0.2274
    /0.1

    so t =-7.408 ?
     
  9. May 14, 2013 #8

    DrClaude

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    Staff: Mentor

    If you don't want to use LaTeX notation, try

    t = (0.5/0.1) ln 0.2274

    The numerical value is correct, but the sign is wrong.
     
  10. May 14, 2013 #9

    I think I made a typo

    t = 7.405

    for future reference is there a 'sticky' guide to using LaTeX notation as I'm currently doing a course and am finding Physics Forums to be of some use!
     
  11. May 14, 2013 #10

    DrClaude

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    Staff: Mentor

  12. May 14, 2013 #11
  13. May 14, 2013 #12

    Mark44

    Staff: Mentor

    No.
    The first two things you should have done, in this order, are:
    1. Divide both sides of the equation by V.
    2. Take the natural log of both sides.

    Starting with θ = Ve^(-Rt/L), step 1 yields
    θ/V = e^(-Rt/L)

    Taking logs of both sides:
    ln(θ/V) = -Rt/L

    Now solve for t.
     
  14. May 6, 2014 #13
    Hi,

    I have the same question in an assignment but get a different answer can anyone advise where I am going wrong please here is my working out.

    θ=Ve^(-Rt)/L

    58 = 255e^(-0.1t)/0.5

    58/255 = e^(-0.1/t)/0.5

    0.2275 = e^(0.1/t)/0.5

    (-0.1t)/0.5 = Log(0.2275)

    -0.1t = 0.5 Log(0.2275)

    t = (0.5Log(0.2275))/-0.1

    t = -0.3215/-0.1

    t = 3.215
     
  15. May 6, 2014 #14
    @Ridley Remember that your taking the natural log, and not log base 10. So

    ##-.1t=.5ln(.2275)##.
     
  16. May 6, 2014 #15
    So I should have wrote.


    -0.1t = 0.5ln(0.2275)

    t = (0.5ln(0.2275))/-0.1

    t = -0.7403 / -0.1

    t = 7.403
     
  17. May 6, 2014 #16
    Yes, that is correct. But do you understand the difference? When you take the logarithm of an exponential you can only "cancel" out the base if and only if the base of your exponent is the same as the base of your logarithm. In other words

    ##log_a(a^n)=n*(log_a(a))=n*(1)=n##

    It is not true however when ##a\neq b## as in the example below

    ##log_b(a^n)=n*(log_b(a))\neq n##
     
    Last edited: May 6, 2014
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