log_{\frac{1}{3}} (\frac{1}{9})\\log_{\frac{1}{3}}(x+15\cdot\frac{1}{9})>log_{\frac{1}{3}}\frac{1}{9}\\\frac{x+15}{9}<\frac{1}{9}\\\frac{x+15}{9}-\frac{1}{9}<0\\\frac{x+15-1}{9}<0\\\frac{x+14}{9}<0\\x+14<0\\x<-14\\x(-\infty;-14)" alt="log_{\frac{1}{3}}(x+15)+log_{\frac{1}{3}}(\frac{1}{9})>log_{\frac{1}{3}} (\frac{1}{9})\\log_{\frac{1}{3}}(x+15\cdot\frac{1}{9})>log_{\frac{1}{3}}\frac{1}{9}\\\frac{x+15}{9}<\frac{1}{9}\\\frac{x+15}{9}-\frac{1}{9}<0\\\frac{x+15-1}{9}<0\\\frac{x+14}{9}<0\\x+14<0\\x<-14\\x(-\infty;-14)" align="absmiddle" class="latex-formula"> Может так, но я не уверенна....