${9^x} = 3\sqrt 3  \Rightarrow {\left( {{3^2}} \right)^x} = 3 \times {3^{\frac{1}{2}}} \Rightarrow {3^{2x}} = {3^{\frac{3}{2}}}$$ \Rightarrow 2x = \frac{3}{2} \Rightarrow x = \frac{{\frac{3}{2}}}{2} = \frac{3}{4}$$\log _2^{\left( {1 - x} \right)} = \log _2^{1 - \frac{3}{4}} = \log _2^{\frac{1}{4}} = \log _2^{{2^{ - 2}}} =  - 2$