Question

b) log2 3 + log2 (x-1)= log2 12
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Answer 1

$log_{2}(3) + log_{2}(x - 1) = log_{2}(12) \\ log_{2}(x - 1) = log_{2}(12) - log_{2}(3) \\ log_{2}(x - 1) = log_{2}( \frac{12}{3} ) \\ \\ log_{2}(x - 1) = log_{2}(4) \\ x - 1 = 4 \\ x = 4 + 1 \\ x = 5$