Solve the following equation for \( x \). \( \log _{12}(x+4)+\log _{12}(x+5)=1 \) \( x=\square \)

Solve the equation \( \log_{12}(x+4)+\log_{12}(x+5)=1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\log_{12}{\left(x+4\right)}+\log_{12}{\left(x+5\right)}=1\) - step1: Find the domain: \(\log_{12}{\left(x+4\right)}+\log_{12}{\left(x+5\right)}=1,x>-4\) - step2: Transform the expression: \(\log_{12}{\left(\left(x+4\right)\left(x+5\right)\right)}=1\) - step3: Convert the logarithm into exponential form: \(\left(x+4\right)\left(x+5\right)=12^{1}\) - step4: Evaluate the power: \(\left(x+4\right)\left(x+5\right)=12\) - step5: Expand the expression: \(x^{2}+9x+20=12\) - step6: Move the expression to the left side: \(x^{2}+9x+20-12=0\) - step7: Subtract the numbers: \(x^{2}+9x+8=0\) - step8: Factor the expression: \(\left(x+1\right)\left(x+8\right)=0\) - step9: Separate into possible cases: \(\begin{align}&x+1=0\\&x+8=0\end{align}\) - step10: Solve the equation: \(\begin{align}&x=-1\\&x=-8\end{align}\) - step11: Check if the solution is in the defined range: \(\begin{align}&x=-1\\&x=-8\end{align},x>-4\) - step12: Find the intersection: \(x=-1\) The solution to the equation \( \log _{12}(x+4)+\log _{12}(x+5)=1 \) is \( x=-1 \).