# Show that x =

• Nov 24th 2010, 02:23 PM
Hellbent
Show that x =
Hi,

My uncertainty lies in the approach to the question. Anyway, I made an attempt.

Given $\displaystyle e^x - e^{-x} = 4$, show that $\displaystyle x = \ln(2+\sqrt{5})$

$\displaystyle \ln e^x - \ln e^{-x} = \ln 4$

$\displaystyle x - (-x) = \ln 4$

$\displaystyle 2x = \ln 4$

$\displaystyle x = \ln 2$
• Nov 24th 2010, 02:26 PM
Plato
$e^x - e^{ - x} = 4\; \equiv \,e^{2x} - 4e^x - 1=0$.
• Nov 24th 2010, 02:32 PM
emakarov
$\displaystyle\ln(e^x/e^{-x})=\ln e^x - \ln e^{-x}$, but $\displaystyle\ln(e^x-e^{-x})\ne\ln e^x - \ln e^{-x}$.
• Nov 24th 2010, 05:11 PM
Hellbent
How did you get this part: $e^{2x} - 4e^x - 1=0?$
Why did you set it as equal to zero?
Is there any other way of doing this?

================================================== ===

$\displaystyle e^{2x} - 4e^x - 1=0$

Let $e^x = y$

$\displaystyle= y^2 - y - 1 = 0$

$\displaystlye= (y - \frac{1}{2})^2 - \frac{5}{4} = 0$

$y = \frac{1}{2}(1 - \sqrt{5})$

$y = \frac{1}{2}(1 + \sqrt{5})$

Resubstituting $y$ for $e^x$:

$e^x = \frac{1}{2}(1 + \sqrt{5})$

$\ln e^x = \ln \frac{1}{2}(1 + \sqrt{5})$

$x = \ln \frac{1}{2}(1 + \sqrt{5})$

OR

$e^x = \ln \frac{1}{2}(1 - \sqrt{5})$

$\ln e^x = \ln \frac{1}{2}(1 - \sqrt{5})$

$x = \ln \frac{1}{2}(1 - \sqrt{5})$

This still isn't it.
• Nov 24th 2010, 05:27 PM
skeeter
$e^x - e^{-x} = 4
$

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

$e^{2x} - 1 = 4e^x$

$e^{2x} - 4e^x - 1 = 0$

quadratic formula (remember $e^x > 0$ for all $x$)

$e^x = 2 + \sqrt{5}$

$x = \ln(2 + \sqrt{5})$
• Nov 24th 2010, 05:55 PM
Hellbent
Thanks.

What's the reason for multiplying throughout by e^x?
No other way of doing this?
• Nov 24th 2010, 05:59 PM
pickslides
Quote:

Originally Posted by Hellbent
What's the reason for multiplying throughout by e^x?
No other way of doing this?

It is done to make the $e^x$ in the numerator

Otherwise $e^x - e^{ - x} - 4 = e^x - \frac{1}{e^{x}}- 4$
• Nov 24th 2010, 06:09 PM
skeeter
Quote:

Originally Posted by Hellbent
Thanks.

What's the reason for multiplying throughout by e^x?
No other way of doing this?

to make the equation quadratic in $e^x$ ... and subsequently easier to solve.
• Nov 25th 2010, 04:43 AM
Hellbent
Quote:

Originally Posted by skeeter
to make the equation quadratic in $e^x$ ... and subsequently easier to solve.

Thanks, I see it now :)