Question

Dealing with logs.
Solve the equation:

e^x + e^-x - 12 = 0

Answer 1

\[e^x + \frac {1}{e^x} - 12 = 0\]\[e^{2x} + 1 - 12e^x = 0 \rightarrow e^{2x} - 12e^x + 1 = 0\]\[e^{x} = \frac {12 \pm \sqrt {144 - 4}}{2}\]\[e^x = 6 \pm \sqrt {35}\]\[x = \ln (6 + \sqrt {35}), x = \ln (6 - \sqrt {35})\]

Answer 2

multiply thru by e^x
e^2x + 1 = 12 e^x
e^2x - 12 e^x + 1 = 0
let y = e^x
y^2 - 12y + 1 = 0
solve for y
and equate roots to e^x

Answer 3

Thanks. I think I understand it! @rogue got the same answer as my test.

Answer 4

Great job =)
I was stumped looking at the problem at first, but I realized the trick was to multiply by e^x =)