Equation Rearrangement

[oiprocs]

I need to isolate k in this equation.

t = (-1/.6k) [ln (1-kv)]

And I can't do it.

/fail? Can anyone help?

 

[esun]

You can't. It's a transcendental equation. If you have a specific value of t and want to solve for a specific value of k, you can use iteration.

 

[JohnCU]

hmm, i took 7 different math courses in college and should know how to do this, but apparently i fail. revoke my degree. i thought maybe i could do something by bringing the (-1/.6k) back into the exponent of the natural log, but i was wrong. i'm trying to remember all the old tricks...hmm..

 

[JohnCU]

oooh nevermind, it's unsolvable so i was partially correct.

 

[oiprocs]

Come on you engineers don't fail me. I'm sure you guys don't do this stuff anymore, but as a college student I must confront it head on, and it's been !hadoiken! all day long.

 

[oiprocs]

Unsolvable? Wtf? What bastards.

 

[JohnCU]

what class is this for? i don't remember hearing about transcendental equations and i went through calc I through 4 then engineering calc.

 

[oiprocs]

Originally posted by: esun
You can't. It's a transcendental equation. If you have a specific value of t and want to solve for a specific value of k, you can use iteration.

Yeah I have values for "t" and for "v", but I wanted to isolate k first. What is iteration?

Btw, t = 20 and v = 6.

 

[oiprocs]

Originally posted by: JohnCU
what class is this for? i don't remember hearing about transcendental equations and i went through calc I through 4 then engineering calc.

Dynamics.

 

[JohnCU]

then you get 20 = (-1/.6*k)*ln(1-6k)
-> 12 = -ln(1-6k)
-> e^12 = 6k-1
k = (e^12+1)/6 ??

 

[JohnCU]

wait that's a hugeamongous number.

 

[oiprocs]

Originally posted by: JohnCU
then you get 20 = (-1/.6*k)*ln(1-6k)
-> 12 = -ln(1-6k)
-> e^12 = 6k-1
k = (e^12+1)/6 ??

You forgot about this other k.

 

[JohnCU]

Originally posted by: Oiprocs

Originally posted by: JohnCU
then you get 20 = (-1/.6*k)*ln(1-6k)
-> 12 = -ln(1-6k)
-> e^12 = 6k-1
k = (e^12+1)/6 ??

You forgot about this other k.

oops. i'm doing this in the dark on a napkin, sorry.

 

[oiprocs]

If it weren't for that extra k, it'd be a cinch, but for some reason it's making the problem ridiculous.

The book says "Solving by trial, k = yadda yadda" but they don't show they went from that equation down to the value of k. Soo, !profit.

 

[oiprocs]

Originally posted by: JohnCU

Originally posted by: Oiprocs

Originally posted by: JohnCU
then you get 20 = (-1/.6*k)*ln(1-6k)
-> 12 = -ln(1-6k)
-> e^12 = 6k-1
k = (e^12+1)/6 ??

You forgot about this other k.

oops. i'm doing this in the dark on a napkin, sorry.

Lol, that reminds of a Beautiful Mind, where Nash writes his equations on window panes.

 

[rgwalt]

Originally posted by: Oiprocs
I need to isolate k in this equation.

t = (-1/.6k) [ln (1-kv)]

And I can't do it.

/fail? Can anyone help?

Rearrange and solve by iteration:

kn = (-20/.6)[ln(1-6k(n-1))]

where n and (n-1) are indicies (not variables). Pick a starting guess for k0, solve for k1. Use k1 to solve for k2, use k2 to solve for k3, repeat until the desired tolerance level is reached. This is called a "fixed point iteration". If the relative difference between successive iterations does not decrease, you may need to start over with a different initial guess. Alternatively, write the equation as:

20 + (1/0.6k)[ln(1-6k)] = 0

and use Newton's method to solve.

R

 

[oiprocs]

Ah sh!t. Well at least I have a starting point. Thank you my friend.

 

[oiprocs]

Well, after 11 guesses I got as close to zero as I think I'll get.

:beer: for you rgwalt.

And you too JohnCU. I guess you gave up and found something better to do. I mean, what could you do when you're in the dark with a napkin? :Q

 

[JohnCU]

Originally posted by: Oiprocs
Well, after 11 guesses I got as close to zero as I think I'll get.

:beer: for you rgwalt.

And you too JohnCU. I guess you gave up and found something better to do. I mean, what could you do when you're in the dark with a napkin? :Q

I simply sat back and waited for someone smarter than I am to help you.

 

[oiprocs]

Haha, you are obviously wiser than the man who proposed the solution.

 

[rgwalt]

Originally posted by: Oiprocs
Well, after 11 guesses I got as close to zero as I think I'll get.

:beer: for you rgwalt.

And you too JohnCU. I guess you gave up and found something better to do. I mean, what could you do when you're in the dark with a napkin? :Q

Glad I could help, math is fun.

R