Solve the equation. $$ \begin{array}{c}\log _{6}(x-2)+\log _{6}(x+3)=2 \ x=[?]\end{array} $$

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Solve the equation $$ \log_{6}(x-2)+\log_{6}(x+3)=2 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\log_{6}{\left(x-2ight)}+\log_{6}{\left(x+3ight)}=2$$
- step1: Find the domain:
  $$\log_{6}{\left(x-2ight)}+\log_{6}{\left(x+3ight)}=2,x>2$$
- step2: Transform the expression:
  $$\log_{6}{\left(\left(x-2ight)\left(x+3ight)ight)}=2$$
- step3: Convert the logarithm into exponential form:
  $$\left(x-2ight)\left(x+3ight)=6^{2}$$
- step4: Evaluate the power:
  $$\left(x-2ight)\left(x+3ight)=36$$
- step5: Expand the expression:
  $$x^{2}+x-6=36$$
- step6: Move the expression to the left side:
  $$x^{2}+x-6-36=0$$
- step7: Subtract the numbers:
  $$x^{2}+x-42=0$$
- step8: Factor the expression:
  $$\left(x-6ight)\left(x+7ight)=0$$
- step9: Separate into possible cases:
  $$\begin{align}&x-6=0\&x+7=0\end{align}$$
- step10: Solve the equation:
  $$\begin{align}&x=6\&x=-7\end{align}$$
- step11: Check if the solution is in the defined range:
  $$\begin{align}&x=6\&x=-7\end{align},x>2$$
- step12: Find the intersection:
  $$x=6$$
The solution to the equation $$ \log_{6}(x-2)+\log_{6}(x+3)=2 $$ is $$ x=6 $$.
$$ x=6 $$

Solve the equation.

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