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

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Solve the equation $$ \log_{12}(x+4)+\log_{12}(x+5)=1 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\log_{12}{\left(x+4ight)}+\log_{12}{\left(x+5ight)}=1$$
- step1: Find the domain:
  $$\log_{12}{\left(x+4ight)}+\log_{12}{\left(x+5ight)}=1,x>-4$$
- step2: Transform the expression:
  $$\log_{12}{\left(\left(x+4ight)\left(x+5ight)ight)}=1$$
- step3: Convert the logarithm into exponential form:
  $$\left(x+4ight)\left(x+5ight)=12^{1}$$
- step4: Evaluate the power:
  $$\left(x+4ight)\left(x+5ight)=12$$
- 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:
  $$\left(x+1ight)\left(x+8ight)=0$$
- 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 $$ x=-1 $$.
$$ x=-1 $$.

Solve the following equation for .

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