$\log _3^{\left( {2x - 1} \right)} =  - 2 \Rightarrow 2x - 1 = {3^{ - 2}}$$ \Rightarrow 2x - 1 = \frac{1}{9} \Rightarrow 2x = \frac{1}{9} + 1$$ \Rightarrow 2x = \frac{{10}}{9} \Rightarrow x = \frac{{\frac{{10}}{9}}}{2} = \frac{{10}}{{18}} = \frac{5}{9}$$\log _2^{9x + 3} = \log _2^{9\left( {\frac{5}{9}} \right) + 3} = \log _2^{5 + 3} = $$\log _2^8 = \log _2^{{2^3}} = 3\log _2^2 = 3$