>>77

t = 3^x + 3^-x ≧ 2 (等号成立は3^x = 3^-xの時、即ちx=0)
として
9^x + 9^-x
= (3^x + 3^-x)^2 - 2
= t^2 - 2

3*(t^2 - 2) - 13t + 16 = 0
3t^2 - 13t + 10 = 0
(3t-10)(t-1) = 0
t≧2より
t = 10/3
3^x + 3^-x = 10/3
3*(3^x)^2 - 10*(3^x) + 3 = 0
{3*(3^x) - 1}*{(3^x) - 3} = 0
3^x = 3,1/3
x = 1,-1