# Help to simplify a difficult equation

1. Oct 19, 2012

### diffrac

Hello,

Could someone please help me simplify the equation:

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

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

so as to get: v = $\frac{(m1-m2)g}{(c1+c2)}$ $\cdot$ [ 1 - exp[$\frac{-(c1+c2)t }{(m1+m2)}$]] (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$\frac{t}{(m1+m2)}$] / [(c1+c2) * exp$\frac{1}{(c1+c2)}$]

Thanks.

2. Oct 19, 2012

### 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.

3. Oct 19, 2012

### diffrac

ok, well i applied e^ to the whole thing which gave me:

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

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

that leaves me with:

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

4. Oct 19, 2012

### 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.

5. Oct 19, 2012

### diffrac

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?

6. Oct 19, 2012

### 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.

7. Oct 19, 2012

### diffrac

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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