Question

Find the value x such that : (See above)​

Answer 1

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Answer 2

$\huge{ \underline{ \sf Solution: }} \\ \\ \\ \sf \rightarrow \quad {3}^{2x - 1} = \frac{1}{ {27}^{x - 3} } \\ \\ \sf \rightarrow \quad {3}^{2x - 1} = {27}^{ - (x - 3)} \qquad \bigg( \because \sf \: \frac{1}{a} = {a}^{ - 1} \: \bigg) \\ \\ \sf \rightarrow \quad {3}^{2x - 1} = {3}^{3(3 - x)} \\ \\ \sf \rightarrow \quad {3}^{2x - 1} = {3}^{9 - 3x} \\ \\ \sf \quad Here, \: The \: bases \: are \: same \: hence \\ \sf \: the \: exponents \: must \: also \: be \: same. \\ \\\sf \rightarrow \quad 2x - 1 = 9 - 3x \\ \\ \sf \rightarrow \quad 2x + 3x = 9 + 1 \\ \\ \sf \rightarrow \quad 5x = 10 \\ \\ \sf \rightarrow \quad x = \frac{ \cancel{10}^{2} }{ \cancel{5}_{1} } \\ \\ \sf \rightarrow \quad \boxed{ \sf \blue{x = 2}}$