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Solve log_2(3x + 7) = 3 + log_2(x – 1), x > 1.

Begin by collecting log_2 terms.log_2(3x + 7) - log_2(x – 1) + = 3 Using the rule of log terms we getlog_2((3x+7)/(x-1)) = 3 (3x+7)/(x-1) = 2³3x+7 = 8(x-1)3x+7-8x+8 = 0 -5x+15 = 0 x= 3

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Solve log_2(3x + 7) = 3 + log_2(x – 1), x > 1.

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