How do you solve #e^x+e^-x=4#?

1 Answer
Aug 22, 2015

#x=ln(2+sqrt(3)) or x=ln(2-sqrt(3))#

Explanation:

Temporarily simplify by letting #y =e^x#

#e^x+e^(-x) =4#
becomes #y+1/y = 4#

Multiplying by #y# and sifting things around:
#color(white)("XXXX")y^2-4y+1=0#
Applying the quadratic formula, we get
#color(white)("XXXX")y =2+sqrt(3)color(white)("XXXX")orcolor(white)("XXXX")y=2-sqrt(3)#

If #y = e^x = 2+sqrt(3)#
#color(white)("XXXX")ln(e^x) = x = ln(2+sqrt(3))#

Similarly, if #y=e^x=2-sqrt(3)#
#color(white)("XXXX")x = ln(2-sqrt(3))#