(4/9)^x + (6/9)^x - 1 = 0

or (2/3)^2x + (2/3)^x - 1 = 0

Put (2/3)^x = y

So, y² + y - 1 = 0

y = (-1 + (5)^1/2) / 2

or, (2/3)^x = (-1 + (5)^1/2) / 2

Taking log on both sides,

x log(2/3) = log (-1 + (5)^1/2) - log(2)

x = [ log (-1 + (5)^1/2) - log(2) ] / [ log(2/3) ] (Ans)............