2\\ x-1>0\\ x>1\\ \log_{\frac{1}{3}}(x-1)>\log_{\frac{1}{3}}(\frac{1}{3})^2\\ x-1<(\frac{1}{3})^2\\ x-1<\frac{1}{9}\\ x<\frac{10}{9}\\ x\in(1,\frac{10}{9}) " alt="\\\log_{\frac{1}{3}}(x-1)>2\\ x-1>0\\ x>1\\ \log_{\frac{1}{3}}(x-1)>\log_{\frac{1}{3}}(\frac{1}{3})^2\\ x-1<(\frac{1}{3})^2\\ x-1<\frac{1}{9}\\ x<\frac{10}{9}\\ x\in(1,\frac{10}{9}) " align="absmiddle" class="latex-formula">