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1. $9^{3} \times {(81)}^{2} \div {(27)}^{3} = {(3)}^{?}$

A. 3

B. 4

C. 5

D. 6

Solution:

\begin{aligned} Let 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}

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