How to Find the Solution to a 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
[tex]cosh(x)= \frac{e^x+ e^{-x}}{2}[/itex]<br /> and<br /> [tex]sinh(x)= \frac{e^x- e^{-x}}{2}[/itex]<br /> <br /> so <br /> $$cosh(x)- 3sinh(x)= \frac{e^x+ e^{-x}}{2}- \frac{3e^x- 3e^{-x}}{2}$$<br /> $$= \frac{-2e^x+ 4e^{-x}}{2}$$<br /> <br /> and the fact that (k, -1) is on the graph means that <br /> $$cosh(k)- 3sinh(k)= \frac{-2e^k+ 4e^{-k}}{2}= -1$$<br /> <br /> Multiplying through by 2 gives<br /> $$-2e^k+ 4e^{-k}= -2$$<br /> <br /> Now, for the step you say you don't understand: Multiply through by $e^k$ to get:<br /> $$-2e^{2k}+ 4= -2e^{k}$$<br /> amd divide by -2 to get<br /> [tex]e^{2k}- 2= e^{k}[/itex]<br /> <br /> Finally, subtract $e^k$ from both sides:<br /> $$e^{2k}- e^k- 2= 0$$[/tex][/tex][/tex]