$7.314 \frac{2-4 \log _{12} 2}{\log _{12}(x+2)}-1=\frac{\log _{6}(8-x)}{\log _{6}(x+2)}$.

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

$7.314 \frac{2-4 \log _{12} 2}{\log _{12}(x+2)}-1=\frac{\log _{6}(8-x)}{\log _{6}(x+2)}$.
## Solution.

$\frac{\log _{12} 144-\log _{12} 16-\log _{12}(x+2)}{\log _{12}(x+2)}=\frac{\log _{6}(8-x)}{\log _{6}(x+2)} \Leftrightarrow \frac{\log _{12} \frac{9}{x+2}}{\log _{12}(x+2)}=\frac{\log _{6}(8-x)}{\log _{6}(x+2)} \Leftrightarrow$

$\Leftrightarrow \log _{x+2}\left(\frac{9}{x+2}\right)=\log _{x+2}(8-x) \Leftrightarrow\left\{\begin{array}{l}x+2 \neq 1, \\ 8-x>0, \\ \frac{9}{x+2}=8-x\end{array} \Leftrightarrow\left\{\begin{array}{l}x \neq-1, \\ x>0, \\ 2-x>0, \\ \lg 2 x(2-x)=\lg \lg p\end{array} \Leftrightarrow\left\{\begin{array}{l}0<x<2, \\ 2 x(2-x)=\lg p\end{array} \Leftrightarrow\left\{\begin{array}{l}0<x<2, \\ 1-\sqrt{1-0.5 \lg p}>0 \Leftrightarrow 1>\sqrt{1-0.5 \lg p} \Leftrightarrow\left\{\begin{array}{l}\lg p>0, \\ \lg p \leq 2\end{array} \Leftrightarrow 10, \\ \lg p \leq 2\end{array} \Leftrightarrow 1<p \leq 100\right.$.

Answer: $x_{1,2}=1 \pm \sqrt{1-0.5 \lg p}$ for $1<p \leq 100$, no solutions for other values of $p$.