Question 526583
2e^x-2e^-x=5 
2e^x-2/e^x=5
let e^x=y. 
2y-2/y=5
2y^2-2=5y
2y^2-5y-2=0
Good so far.
*[invoke solve_quadratic_equation 2,-5,-2]
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e^x = the solutions
x = ln of the positive solution
x =~ ln(2.85)
x =~ 1.04732