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Unregistered
11th June 2006, 03:38 PM
how do you solve 3e^y + 5e^-y = 16 ???
please help

ExamSolutions
11th June 2006, 06:27 PM
3e^y + 5e^-y = 16

multiply both sides by e^y
3[(e^y)^2]+5=16e^y
let x=e^y
3x-16x+5=0
(3x-1)(x-5)=0
x=1/3 or x=5
e^y = 1/3 or e^y = 5
y = ln(1/3) or y=ln(5)