Solve \(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\).
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Correct Answer: Option B
Explanation:
\(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\)
Recall, \( 1 = \log_{2}2\), so
\(\log_{2}(12x - 10) = \log_{2}2 + \log_{2}(4x + 3)\)
= \(\log_{2}(12x - 10) = \log_{2}(2(4x + 3))\)
\(\implies (12x - 10) = 2(4x + 3) \therefore 12x - 10 = 8x + 6\)
\(12x - 8x = 4x = 6 + 10 = 16 \implies x = 4\)
\(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\)
Recall, \( 1 = \log_{2}2\), so
\(\log_{2}(12x - 10) = \log_{2}2 + \log_{2}(4x + 3)\)
= \(\log_{2}(12x - 10) = \log_{2}(2(4x + 3))\)
\(\implies (12x - 10) = 2(4x + 3) \therefore 12x - 10 = 8x + 6\)
\(12x - 8x = 4x = 6 + 10 = 16 \implies x = 4\)