0} \atop {\frac{2}{x}>0}} \right. \; \left \{ {{x>-1} \atop {x>0}} \right.\; \to \; x>0\\\\\frac{x+1}{10}=\frac{2}{x}\; \; \to \; \; x^2+x=20\; ,\\\\x^2+x-20=0\; ,\; \; x_1=-5\; ,\; x_2=4>0\; \; (teorema\; Vieta)\\\\Otvet:\; \; x=4\; ." alt="2)\; \; log_5(\frac{x+1}{10})=log_5(\frac{2}{x})\; \; ,\; \; ODZ:\; \; \left \{ {{\frac{x+1}{10}>0} \atop {\frac{2}{x}>0}} \right. \; \left \{ {{x>-1} \atop {x>0}} \right.\; \to \; x>0\\\\\frac{x+1}{10}=\frac{2}{x}\; \; \to \; \; x^2+x=20\; ,\\\\x^2+x-20=0\; ,\; \; x_1=-5\; ,\; x_2=4>0\; \; (teorema\; Vieta)\\\\Otvet:\; \; x=4\; ." align="absmiddle" class="latex-formula">