# [SOLVED] exponential e problem, solve for roots

Printable View

• Sep 24th 2009, 03:39 PM
tkdiamond08
[SOLVED] exponential e problem, solve for roots
Hey all, I seem to be having some trouble finding the solutions (roots) to this equation:

0=210-10(e^(x/30) + e^(-x/30))

I got it down to this lol:

21=e^(x/30) + e^(-x/30)

and it seems like it should be easy, just I can't seem to simplify it down enough to be able to take the natural log from both sides :(

Any help would be greatly appreciated,

thanks!
• Sep 24th 2009, 03:53 PM
mr fantastic
Quote:

Originally Posted by tkdiamond08
Hey all, I seem to be having some trouble finding the solutions (roots) to this equation:

0=210-10(e^(x/30) + e^(-x/30))

I got it down to this lol:

21=e^(x/30) + e^(-x/30)

and it seems like it should be easy, just I can't seem to simplify it down enough to be able to take the natural log from both sides :http://mathhelpforum.com/pre-calculus/104124-solved-exponential-e-problem-solve-roots.html#post370919\" rel=\"nofollow\">
Hey all, I seem to be having some trouble finding the solutions (roots) to this equation:

0=210-10(e^(x/30) + e^(-x/30))

I got it down to this lol:

21=e^(x/30) + e^(-x/30)

and it seems like it should be easy, just I can't seem to simplify it down enough to be able to take the natural log from both sides :(

Any help would be greatly appreciated,

thanks!

let $u = \frac{x}{30}$

$21 = e^u + e^{-u}$

multiply every term by $e^u$ ...

$21e^u = e^{2u} + 1$

$0 = (e^{u})^2 - 21e^u + 1$

quadratic formula ...

$e^u = \frac{21 \pm \sqrt{(-21)^2 - 4}}{2}$

$e^u = \frac{21 \pm \sqrt{437}}{2}$

$u = \ln\left(\frac{21 \pm \sqrt{437}}{2}\right)$

$x = 30\ln\left(\frac{21 \pm \sqrt{437}}{2}\right)$
• Sep 24th 2009, 04:14 PM
tkdiamond08
Thanks guys!

Oh and sorry for initially posting it in the wrong forum.