# stuck on "e" and "ln" stuff

• Jun 17th 2006, 02:23 PM
Dalau
stuck on "e" and "ln" stuff
I'm stuck on how to work with "e" and "ln" in certain ways. This is for a calculus project about track and field statistics and growth constants.

I'll post the longer stuff later in the post.

edit: you don't have to read the long part. offering help on the short part right below will be fine. thanks =)
So here's the quick end of things:

Known variables:
Vn1, Vo1, Vn2, Vo2, t1, t2

The unknown variable is A.

how can I simplify this. I'm basically trying to get rid of the "ln".

ln[(Vn1-A)/(Vo1-A)] = (t1/t2)*ln[(Vn2-A)/(Vo2-A)

I tried doing "e" raised to everything on both sides, so it looks like this:

e^{ln[(Vn1-A)/(Vo1-A)]} = e^{(t1/t2)*ln[(Vn2-A)/(Vo2-A)}

And that was when I got stuck. I know I can't split the right side into two parts. For example, I can't do:
e^{(t1/t2)} * e^{ln[(Vn2-A)/(Vo2-A)}

What do I do then?

OK, here's the longer part, just to check if I did anything wrong to begin with. The original problem looks like this. It's two equations to begin with. Variable "k" cancels out later, as I'll show.

Vn1 = A + (Vo1 - A)*e^(k*t1)

and

Vn2 = A + (Vo2 - A)*e^(k*t2)

Now I simplify each equation, and get this:

[(Vn1-A)/(Vo1-A)] = e^(k*t1)

and

[(Vn2-A)/(Vo2-A)] = e^(k*t2)

From here, I multiplied everything by ln, and got:

ln[(Vn1-A)/(Vo1-A)] = k*t1

and

ln[(Vn2-A)/(Vo2-A)] = k*t2

When I do systems of equations, I can cancel out the "k", so it becomes:

( ln[(Vn1-A)/(Vo1-A)] / ln[(Vn2-A)/(Vo2-A)] ) = (t1/t2)

After that, I can either rearrange the equation and get the equation I posted at the beginning of the post, or I can rearrange and get this:

ln[(Vn1-A)/(Vo1-A)] / t1 = ln[(Vn2-A)/(Vo2-A)] / t2

I'm not sure which way works better, but either way, i'm trying to still find "A".

Any way you could help would be appreciated. Thanks,
-Dan
• Jun 17th 2006, 06:59 PM
ThePerfectHacker
From the equations system you have,

$\left\{\begin{array}{c}
V_{n1} = A + (V_{01} - A)e^{kt_1}\\
V_{n2} = A + (V_{02} - A)e^{kt_2}
$

Why would you ever combine them?!?
The solution is right there.
Simply work with the first one (or second).
$V_{n1}=A+V_{01}e^{kt_1}-Ae^{kt_1}$
Thus,
$V_{n1}-V_{01}e^{kt_1}=A-Ae^{kt_1}$
Thus,
$V_{n1}-V_{01}e^{kt_1}=A(1-e^{kt_1})$
Thus,
$A=\frac{V_{n1}-V_{01}e^{kt_1}}{1-e^{kt_1}}$
• Jun 17th 2006, 07:02 PM
Dalau
oops. darn.

i forgot to say that "k" is also an unknown, which is why i combined equations to try to cancel it out. my appologies. that was my fault.
• Jun 17th 2006, 07:08 PM
ThePerfectHacker
Quote:

Originally Posted by Dalau
oops. darn.

i forgot to say that "k" is also an unknown, which is why i combined equations to try to cancel it out. my appologies. that was my fault.

I did not try it because I am going to sleep soon. But it does you difficult and perhaps no elementary method.

It does, however, look like some faulty attempt to solve a system of linear-homogenous diffrencial equations, does it? If it is may you display these equations I might be able to solve them.
• Jun 17th 2006, 07:14 PM
Dalau
so these are the two original equations:

$\left\{\begin{array}{c}
V_{n1} = A + (V_{01} - A)e^{kt_1}\\
V_{n2} = A + (V_{02} - A)e^{kt_2}
$

They're heating/cooling equations used in calculus 2, but I'm using it for a slightly different situation. The "V"s represent the velocity of the world record running pace (m/s) for a certain running event. "A" is the unknown maximum world record velocity in that event over the course of an infinite number of years. 'k" is the growth constant. "t" is the time period between two chosen world records. it can be any of the world record over time. So basically, what i'm doing in this project is trying to find what velocity the world record pace is approaching over time.
• Jun 17th 2006, 07:29 PM
Dalau
quick question:
is there a way to solve ln(x-y)?
• Jun 17th 2006, 07:35 PM
ThePerfectHacker
Quote:

Originally Posted by Dalau
quick question:
is there a way to solve ln(x-y)?

If you mean,
$\ln (x-y)=y$
No.
However you can bring it Lambert W function form (but is is inelementary).