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

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let {15.9}^{3} \times {(81)}^{2} \div {(27)}^{3} = {(3)}^{x} then \\ \Rightarrow {(3)}^{x} = \frac{{(3^{2})}^{3} \times {(3^{4})}^{2}}{{(3^{3})}^{3}} \\ \Rightarrow {(3)}^{x} = \frac{3^{(2 \times 3)} \times 3^{(4 \times 2)}}{3^{(3 \times 3)}} \\ \Rightarrow {(3)}^{x} = \frac{3^{6} \times 3^{8}}{3^{9}} \\ \Rightarrow {(3)}^{x} = \frac{3^{(6 + 8)}}{3^{9}} \\ \Rightarrow {(3)}^{x} = \frac{3^{14}}{3^{9}} \\ \Rightarrow {(3)}^{x} = 3^{(14 - 9)} \\ \Rightarrow {(3)}^{x} = 3^{5} \\ \Rightarrow {(3)}^{x} = 5 \end{aligned}

$\eqalign{ & {\text{Let}}\,{15.9^3} \times {\left( {81} \right)^2} \div {\left( {27} \right)^3} = {\left( 3 \right)^x}{\text{then}} \cr & \Rightarrow {\left( 3 \right)^x}{\text{ = }}\frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {{3^4}} \right)}^2}}}{{{{\left( {{3^3}} \right)}^3}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {2 \times 3} \right)}} \times {3^{\left( {4 \times 2} \right)}}}}{{{3^{\left( {3 \times 3} \right)}}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^6} \times {3^8}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {6 + 8} \right)}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{14}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^{\left( {14 - 9} \right)}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^5} \cr & \Rightarrow {\left( 3 \right)^x} = 5 \cr}$