Let f(x)= π^2 and g(x) = e^x. Then solve the following equation and
give answer(s) in exact form:
f(x-1)-2∙g(2x-1)=0
f(x)= π^2
f(x-1) = π^2, since there is no x in the expression, f(x) = π^2 for all x's
g(x) = e^x
g(2x-1) = e^(2x-1)
f(x-1)-2∙g(2x-1)=0
π^2 - 2 e^(2x-1) = 0
2 e^(2x-1) = π^2
e^(2x-1) = π^2/2
2x - 1 = ln(π^2/2) = 2lnπ - ln2
2x = 2lnπ - ln2 + 1
x = lnπ - (1/2)ln2 + 1/2