ОДЗ: основание log должно быть >0 и не 1, а агумент >0, то есть
0\; ,\; x-5\ne 1} \atop {(x-5)^2>0}} \right. \; \left \{ {{x>5\; ,\; x\ne 6} \atop {x-5\ne 0}} \right. \; \left \{ {{x>5\; ,\; x\ne 6} \atop {x\ne 5}} \right. \; \; \to \\\\\underline {\; x\in (5,6)\cup (6,+\infty )\; }\\\\49=(x-5)^2\\\\x^2-10x-24=0\\\\x_1=-2\notin ODZ\; \; \; ,\; \; x_2=12\in ODZ\\\\\underline {Otvet:\; \; x=12\; .}" alt="log_{x-5}49=log_{x-5}(x-5)^2\\\\ODZ:\; \; \left \{ {{x-5>0\; ,\; x-5\ne 1} \atop {(x-5)^2>0}} \right. \; \left \{ {{x>5\; ,\; x\ne 6} \atop {x-5\ne 0}} \right. \; \left \{ {{x>5\; ,\; x\ne 6} \atop {x\ne 5}} \right. \; \; \to \\\\\underline {\; x\in (5,6)\cup (6,+\infty )\; }\\\\49=(x-5)^2\\\\x^2-10x-24=0\\\\x_1=-2\notin ODZ\; \; \; ,\; \; x_2=12\in ODZ\\\\\underline {Otvet:\; \; x=12\; .}" align="absmiddle" class="latex-formula">