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Help to simplify a difficult equation

  1. Oct 19, 2012 #1
    Hello,

    Could someone please help me simplify the equation:

    [itex]\frac{-1}{(c1+c2)}[/itex][itex] [/itex] [itex]\cdot[/itex] ln [(m1-m2)g-(c1+c2)v] = [itex]\frac{t}{(m1+m2)}[/itex] - [itex]\frac{1}{(c1+c2)}[/itex] [itex]\cdot[/itex] ln[(m1-m2)g]

    (where c1, c2, m1, m2, g, are constants)

    so as to get: v = [itex]\frac{(m1-m2)g}{(c1+c2)}[/itex] [itex]\cdot[/itex] [ 1 - exp[[itex]\frac{-(c1+c2)t }{(m1+m2)}[/itex]]] (teacher's solution)?


    I tried several ways to solve the equation for v (applying e^ to it), but all i can seem to get is:

    v = [exp[itex]\frac{t}{(m1+m2)}[/itex]] / [(c1+c2) * exp[itex]\frac{1}{(c1+c2)}[/itex]]

    Thanks.
     
  2. jcsd
  3. Oct 19, 2012 #2

    Mark44

    Staff: Mentor

    Was that the first thing you did? If so, that's not the best approach. The first thing to do would be to multiply both sides by -(c1 + c2).

    In any case, show us what you did.
     
  4. Oct 19, 2012 #3
    ok, well i applied e^ to the whole thing which gave me:

    -(m1-m2)g + (c1+c2)v e^[itex]\frac{1}{c1+c2}[/itex] = e^[itex]\frac{t}{m1+m2}[/itex] -(m1-m2)g e^[itex]\frac{1}{c1+c2}[/itex].

    I divided that by (m1-m2)g, and then 'simplified' that by eliminating the -e^([itex]\frac{1}{c1+c2}[/itex]) left on both sides of the equation.

    that leaves me with:

    e^[itex]\frac{1}{c1+c2}[/itex] [(c1+c2)v] / (m1-m2)g = e^[itex]\frac{t}{m1+m2}[/itex] / (m1-m2)g.

    i then isolated the v on the left of the equation, and on the right simplified the (m1-m2)g, which finally gives me the result that i put on the first post...

    ok so after i multiply both sides by -(c1+c2), i have to raise everything to e^ to get rid of the natural logarithms, right? thanks for your help
     
  5. Oct 19, 2012 #4

    Mark44

    Staff: Mentor

    Your mistake is above. You are essentially saying that ea*b = ea * eb, which is not true.

    I followed my own advice (by first multiplying both sides by -(c1 + c2) ) and got the teacher's solution.
     
  6. Oct 19, 2012 #5
    Oh right...classic mistake. Although i did try it again and got the wrong answer.
    Could i ask you to show me your steps to the result?
     
  7. Oct 19, 2012 #6

    Mark44

    Staff: Mentor

    It's a lot of stuff to write with all the subscripts and exponents, so I'll only do a couple of steps.

    Multiply both sides by -(c1 + c2):

    ln[(m1 - m2)g - (c1 + c2)v] = -(c1 + c2)*t/(m1 + m2) + ln[(m1 - m2)g]

    Exponentiate both sides (i.e. write each side as an exponent on e):

    (m1 - m2)g = eln[(m1 - m2)g] - (c1 + c2)t/(m1 + m2)

    I'm guessing that if you got this far you might have made a mistake in the next step, by saying that ea - b = ea - eb -- not true. If you made this mistake, take a few minutes and review the properties of exponents.
     
  8. Oct 19, 2012 #7
    Hey great! I finally got it right this time.
    It was one or two of those exponent rules that i tripped over, but i think i got it cleared up now.

    So thank you so much for your help and details Mark44!
     
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