Given that \(\log_{3}(x - y) = 1\) and \(\log_{3}(2x + y) = 2\), find the value of x.
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Correct Answer: Option D
Explanation:
\(\log_{3}(x - y) = 1 \implies x - y = 3^{1} = 3 .... (1)\)
\(\log_{3}(2x + y) = 2 \implies 2x + y = 3^{2} = 9 ..... (2)\)
From (1), y = x - 3
From (2), y = 9 - 2x
\(\implies 9 - 2x = x - 3\)
\(9 + 3 = x + 2x = 3x\)
\(x = 4\)
\(\log_{3}(x - y) = 1 \implies x - y = 3^{1} = 3 .... (1)\)
\(\log_{3}(2x + y) = 2 \implies 2x + y = 3^{2} = 9 ..... (2)\)
From (1), y = x - 3
From (2), y = 9 - 2x
\(\implies 9 - 2x = x - 3\)
\(9 + 3 = x + 2x = 3x\)
\(x = 4\)