let y = 3^x , then 9^x = (3^2)^x = (3^x)^2 = y^2, so we can write the equation as
y^2 - 10y + 9 = 0
(y - 9)(y - 1) = 0
giving y = 9 or y = 1
Therefore 3^x = 9 giving x = 2
and 3^x = 1 giving x = 0
so the solution set is {0, 2}
Note - I have assumed that the 'x' after the 10 means 'multiply'. Is that correct?
