How do you solve #(e^x+e^-x)/2=3#?

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Answer 1 · 2016-06-07T03:07:20

#x≈1.82#

From the given equation, we derive that

#e^x - e^-x = 6#.

Note that #e^-x = 1/(e^x)# and the expression can be simplfied as

#e^x - 1/(e^x) = 6#
#(e^x)^2 - 6(e^x) - 1 = 0#

#e^x = (6 pm sqrt(36-4(-1)))/(2)= (6 pm sqrt(40))/(2)= (6 pm 2sqrt(10))/(2)=3 pm sqrt(10)#

Thus, #e^x = 3+sqrt(10), x = ln(3+sqrt(10)) = 1.8184 ≈ 1.82# (3sf)

or

#e^x = 3-sqrt(10) < 0# (no solutions)