Hyperbolic Graph Problem

[trojsi]

Hi,
please find attached the problem and the short and sweet Answer.
I can't understand the last step of the answer.

[HallsofIvy]

Given that (k, -1) is on the graph of y= cosh(x)- 3sinh(x), show that
e^{2k}- e^k- 2= 0

First the part you say you understand, but I'll write it out so others can follow:

By definition
cosh(x)= \frac{e^x+ e^{-x}}{2}[/itex]
and
[tex]sinh(x)= \frac{e^x- e^{-x}}{2}[/itex]

so
[tex]cosh(x)- 3sinh(x)= \frac{e^x+ e^{-x}}{2}- \frac{3e^x- 3e^{-x}}{2}
= \frac{-2e^x+ 4e^{-x}}{2}

and the fact that (k, -1) is on the graph means that
cosh(k)- 3sinh(k)= \frac{-2e^k+ 4e^{-k}}{2}= -1

Multiplying through by 2 gives
-2e^k+ 4e^{-k}= -2

Now, for the step you say you don't understand: Multiply through by e^k to get:
-2e^{2k}+ 4= -2e^{k}
amd divide by -2 to get
[tex]e^{2k}- 2= e^{k}[/itex]

Finally, subtract e^k from both sides:
e^{2k}- e^k- 2= 0