log 2 (9-2^x) = 3-x ......in exponential form, we have
2^(3-x) = 9 - 2^x
2^(3 - x) + 2^x = 9
2^3 / 2^x + 2^x = 9
2^3 + 2^(2x) = 9^(x)
2^(2x) - 9^(x) + 8 = 0
(2^x - 8) ( 2^x - 1) = 0
So we have that either
2^x - 8 = 0 or 2^x - 1 = 0
2^x = 8 2^x = 1
x = 3 x = 0
