1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Help to simplify a difficult equation

  1. Oct 19, 2012 #1

    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]]

  2. jcsd
  3. Oct 19, 2012 #2


    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


    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


    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook