$${9^3} \times {\left( {81} \right)^2} \div {\left( {27} \right)^3} = {\left( 3 \right)^?}$$

Solution (By Examveda Team)

$$\eqalign{ & {\text{Let}}\,{9^3} \times {\left( {81} \right)^2} \div {\left( {27} \right)^3} = {\left( 3 \right)^x}{\text{then}} \cr & \Rightarrow {\left( 3 \right)^x}{\text{ = }}\frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {{3^4}} \right)}^2}}}{{{{\left( {{3^3}} \right)}^3}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {2 \times 3} \right)}} \times {3^{\left( {4 \times 2} \right)}}}}{{{3^{\left( {3 \times 3} \right)}}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^6} \times {3^8}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {6 + 8} \right)}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{14}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^{\left( {14 - 9} \right)}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^5} \cr & \Rightarrow {\left( 3 \right)^x} = 5 \cr} $$