Solve

9^x-10x3^x+9=0

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?