Solving e^x + e^-x = 6

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Philoctetes3
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I just want to make sure that I have this in the correct order. The book I have is very unclear.

e^x + e^-x = 6
e^x + 1/e^x = 6
e^2x + 1 = 6e^x
e^2x/e^x + 1 = 6
x = ln 6

Thanks.
 

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You've made a mistake. The fourth line doesn't follow from the third. It looks like you tried to divide by e^x (which is actually just undoing what you did in going from line 2 to 3), but you forgot to divide the 1 by e^x.

What you should do is rearrange your third line so that it looks like

e^(2x) - 6e^x + 1 = 0.

Does this look like a familar sort of equation? (If not, let y = e^x. Then does it look familar?)
 
Thanks a lot. I thought something looked off. The thing that worried me is that the answer that I got is quite approximate, and I wanted to make sure that I was doing the process correctly, which I wasn't.

Then after this I would simply do the U form of the quadratic, correct?
 
Yes, the equation is quadratic in form, so a substitution will make it a quadratic, which you can solve using the quadratic formula.