The equation can be rewritten as 3^x+3(3^(-x))<4 and by multiplying by 3^x (which we are allowed to do so since the number 3^x is positive for all values of x) we obtain the equation 3^(2x)-4(3^x)+3<0. If we set y=3^x then we obtain the quadratic equation y^2-4y+3<0. The two roots are 1 and 3. Thus, in order for y^2-4y+3 to be negative we must have 1<y=e^x<3 which implies that 0<x<1 since (1=3^0 and 3=3^1).
