${\log }_2(x+6)+{\log }_2(x+5)=1$

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 $(0,\mathrm{\infty })$.

Step 8 ：$\boxed{{\displaystyle \text{Therefore, the original equation }{\log }_2(x+6)+{\log }_2(x+5)=1\text{ has no solution.}}}$