Help to simplify a difficult equation

[diffrac]

Hello,

Could someone please help me simplify the equation:

$\frac{-1}{(c1+c2)}$[itex][/itex] $\cdot$ 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.

 

[Mark44]

Hello,

Could someone please help me simplify the equation:

$\frac{-1}{(c1+c2)}$[itex][/itex] $\cdot$ 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),

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.

but all i can seem to get is:

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

Thanks.

 

[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

 

[Mark44]

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^$\frac{1}{c1+c2}$.

Your mistake is above. You are essentially saying that e^a*b = e^a * e^b, which is not true.

I followed my own advice (by first multiplying both sides by -(c₁ + c₂) ) and got the teacher's solution.

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

 

[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?

 

[Mark44]

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 = e^{ln[(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 e^{a - b} = e^a - e^b -- not true. If you made this mistake, take a few minutes and review the properties of exponents.

 

[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!