Ответ:
0\\x-1>0\end{array}\right\ \ ,\ \underline {\underline {x>3}}\ ,\\\\\\\dfrac{x-3}{x-1}=2\ \ ,\ \ \dfrac{x-3}{x-1}-2=0\ ,\ \ \dfrac{x-3-2x+2}{x-1}=0\ \ ,\ \ \dfrac{-x-1}{x-1}=0\ \Rightarrow \\\\\\-x-1=0\\\\x=-1" alt="log_7(x-3)-log_7(x-1)=log_72\ \ ,\ \ \ ODZ:\ \left\{\begin{array}{ccc}x-3>0\\x-1>0\end{array}\right\ \ ,\ \underline {\underline {x>3}}\ ,\\\\\\\dfrac{x-3}{x-1}=2\ \ ,\ \ \dfrac{x-3}{x-1}-2=0\ ,\ \ \dfrac{x-3-2x+2}{x-1}=0\ \ ,\ \ \dfrac{-x-1}{x-1}=0\ \Rightarrow \\\\\\-x-1=0\\\\x=-1" align="absmiddle" class="latex-formula">