$\log _{2}(x+6)+\log _{2}(x+5)=1$

Step 1 ：\(\log _{2}(x+6)+\log _{2}(x+5)=1\) becomes \(\log _{2}[(x+6)(x+5)]=1\) using the property of logarithms \(\log_b(m) + \log_b(n) = \log_b(mn)\).

Step 2 ：\(\log _{2}[(x+6)(x+5)]=1\) becomes \(2^1 = (x+6)(x+5)\) by converting the logarithmic equation to an exponential equation.

Step 3 ：\(2 = (x+6)(x+5)\) simplifies to \(2 = x^2 + 11x + 30\).

Step 4 ：\(2 = x^2 + 11x + 30\) rearranges to \(x^2 + 11x + 30 - 2 = 0\), which simplifies to \(x^2 + 11x + 28 = 0\).

Step 5 ：\(x^2 + 11x + 28 = 0\) factors to \((x+4)(x+7) = 0\).

Step 6 ：Setting each factor equal to zero gives \(x = -4\) and \(x = -7\).

Step 7 ：Checking the solutions in the original equation, we find that both \(x = -4\) and \(x = -7\) are undefined in the original equation because the domain of the logarithm function is \((0, \infty)\).

Step 8 ：\(\boxed{\text{Therefore, the original equation } \log _{2}(x+6)+\log _{2}(x+5)=1 \text{ has no solution.}}\)