(3.) \( 6 \cdot \log _{2} 2=\log _{2} 8+\log _{2}(a-2) \)

To solve the equation \( 6 \cdot \log _{2} 2=\log _{2} 8+\log _{2}(a-2) \), we start by simplifying both sides. We know that \( \log_{2} 2 = 1 \), so the left side simplifies to \( 6 \). Now, let's simplify the right side. Using the property of logarithms that states \( \log_{b} m + \log_{b} n = \log_{b} (mn) \), we can rewrite the right side: \[ \log_{2} 8 + \log_{2}(a-2) = \log_{2}(8(a-2)) \] Now we can express 8 as \( 2^3 \), so \( \log_{2} 8 = 3 \). Thus, we have: \[ \log_{2}(8(a-2)) = 3 + \log_{2}(a-2) \] Now, we set up the equation: \[ 6 = 3 + \log_{2}(a-2) \] Subtracting 3 from both sides gives: \[ 3 = \log_{2}(a-2) \] Exponentiating both sides with base 2 results in: \[ a - 2 = 2^3 \] This simplifies to: \[ a - 2 = 8 \implies a = 10 \] Thus, the solution is \( a = 10 \).