$\log _{2}(x)+\log _{2}(x-12)=6$

Step 1 ：Given the logarithmic equation \(\log _{2}(x)+\log _{2}(x-12)=6\)

Step 2 ：Combine the two logarithms into one using the property of logarithms \(\log_b(m) + \log_b(n) = \log_b(mn)\), which simplifies the equation to \(\log _{2}(x(x-12))=6\)

Step 3 ：Convert the logarithmic equation into an exponential equation using the property of logarithms \(b^y = x\) if and only if \(\log_b(x) = y\), which gives us \(x(x-12)=2^6\)

Step 4 ：Solve the quadratic equation to find the value of x, which gives us the solutions \(x = -4\) and \(x = 16\)

Step 5 ：However, since the logarithm is not defined for negative numbers and zero, the only valid solution is \(x = 16\)

Step 6 ：Final Answer: The solution to the equation \(\log _{2}(x)+\log _{2}(x-12)=6\) is \(\boxed{16}\)