How do you solve #e^x + e^(-x) = 3#?

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Answer 1 · 2016-01-01T18:02:37

Express as a quadratic in #t = e^x#, solve and take logs to find:

#x = ln((3+-sqrt(5))/2) =+-ln((3+sqrt(5))/2)#

Let #t = e^x#.

Then the equation becomes:

#t + 1/t = 3#

Multiplying both sides by #t# we get:

#t^2+1 = 3t#

Subtract #3t# from both sides to get:

#t^2-3t+1 = 0#

#t = (3+-sqrt(5))/2#

Note that due to the symmetry of the equation #t+1/t = 3# in #t# and #1/t#, these two values are actually reciprocals of one another.

Now #t = e^x#, so:

#e^x = (3+-sqrt(5))/2#

Taking natural logs of both sides we find:

#x = ln((3+-sqrt(5))/2) =+-ln((3+sqrt(5))/2)#