Math problem

[PowerMacG5]

Originally posted by: lowfatbaconboy
ti-89

 

[vtqanh]

lnx is an increasing function while -3x is a decreasing function, if you can find one root, that's gonna be the only one.
And the root of this equation cannot be found alrebraically. Just use Matlab or TI-89 to find an approximate of the root, and then use derivatives to prove that one is increasing while the other one is decreasing, that will do it

 

[Lizardman]

take the derivative of both sides. That will do it.

so you get
1/2*d(ln(x))/dx = -3*dx/dx
so you have
1/2 *1/x=-3

I think you can take care of the rest
edit: I think that is right. Are there any examples in your book.

 

[vtqanh]

Originally posted by: Lizardman
take the derivative of both sides. That will do it.

so you get
1/2*d(ln(x))/dx = -3*dx/dx
so you have
1/2 *1/x=-3

I think you can take care of the rest
edit: I think that is right. Are there any examples in your book.

huh?

 

[DrPizza]

Originally posted by: Lizardman
take the derivative of both sides. That will do it.

so you get
1/2*d(ln(x))/dx = -3*dx/dx
so you have
1/2 *1/x=-3

I think you can take care of the rest
edit: I think that is right. Are there any examples in your book.

No, that won't do it. Easy counter example just to show how incorrect your method is..
x squared = 4. Take the derivative of both sides. 2x = 0. "I think you can take care of the rest" and see that your method has a flaw in it somewhere..... (big flaw!)

 

[eLiu]

Originally posted by: Lizardman
take the derivative of both sides. That will do it.

so you get
1/2*d(ln(x))/dx = -3*dx/dx
so you have
1/2 *1/x=-3

I think you can take care of the rest
edit: I think that is right. Are there any examples in your book.

haha...dude you cheated at math!

ok but seriously...newton raphson method is probably the easiest. If you don't know calculus...fixed point iteration (significantly slower for functions that don't converge rapidly) is the next one. The W(x) function mentioned in the "Ask Dr. Math" link is also good...but you use that...means you have a computer available...in which case it'd be easier to just say solve(function,x); or some such. There are also a bunch of other methods for solving equations like that...but this is all I know, lol.

edit: fixed point doesn't always work either. There's a certain condition that the function has to satisfy in order for fixed point to work. It had somethign to do with the function's behavior within a certain range of the root (something with a derivative), but I completely forgot what it is.