[SOLVED] exponential e problem, solve for roots

• Sep 24th 2009, 02:39 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, 02:53 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 :(

Any help would be greatly appreciated,

thanks!

Let $\displaystyle w = e^{x/30}$: $\displaystyle 21 = w + \frac{1}{w}$.

Re-arrange this into a quadratic equation in w, solve for w and hence solve for x.
• Sep 24th 2009, 02:54 PM
skeeter
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 :(

Any help would be greatly appreciated,

thanks!

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

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

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

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

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

$\displaystyle e^u = \frac{21 \pm \sqrt{(-21)^2 - 4}}{2}$
$\displaystyle e^u = \frac{21 \pm \sqrt{437}}{2}$
$\displaystyle u = \ln\left(\frac{21 \pm \sqrt{437}}{2}\right)$
$\displaystyle x = 30\ln\left(\frac{21 \pm \sqrt{437}}{2}\right)$