If 27\(^{x + 2}\) \(\div\) 9\(^{x + 1}\) = 3\(^{2x}\), find x.
\(27^{x + 2} \div 9^{x + 1} = 3^{2x}\)
\(3^{3(x + 2)} \div 3^{2(x + 1)} = 3^{2x}\)
\(3^{3x + 6} \div 3^{2x + 2} = 3^{2x}\)
\(\therefore (3x + 6) - (2x + 2) = 2x\)
\(x + 4 = 2x \implies x = 4\)
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