0} \atop {2^{x+1}-2>0}} \right. ; \left \{ {{2^x>2^0} \atop {2(2^x-1)>0}} \right;\ x>0\\\log_2(2^x-1)\cdot\log_{2^{-1}}[2(2^x-1)]>-2;\\-\log_2(2^x-1)(\log_22+\log_2(2^x-1))>-2;\ \log_2(2^x-1)=t;\\ " alt=" \left \{ {{2^x-1>0} \atop {2^{x+1}-2>0}} \right. ; \left \{ {{2^x>2^0} \atop {2(2^x-1)>0}} \right;\ x>0\\\log_2(2^x-1)\cdot\log_{2^{-1}}[2(2^x-1)]>-2;\\-\log_2(2^x-1)(\log_22+\log_2(2^x-1))>-2;\ \log_2(2^x-1)=t;\\ " align="absmiddle" class="latex-formula">
-2} \atop {\log_2(2^x-1)<1}} \right. ; \left \{ {{2^x-1>\frac{1}{4}} \atop {2^x-1<2}} \right. ; \left \{ {{2^x>\frac{5}{4}} \atop {2^x<3}} \right.; \left \{ {{x>\log_2\frac{5}{4}} \atop {x<\log_23}} \right. ; x\in(\log_25-2;\log_23) " alt=" t(1+t)<2;\ t^2+t-2<0;\ (t+2)(t-1)<0;\ t\in(-2;1);\ \\\left \{ {{\log_2(2^x-1)>-2} \atop {\log_2(2^x-1)<1}} \right. ; \left \{ {{2^x-1>\frac{1}{4}} \atop {2^x-1<2}} \right. ; \left \{ {{2^x>\frac{5}{4}} \atop {2^x<3}} \right.; \left \{ {{x>\log_2\frac{5}{4}} \atop {x<\log_23}} \right. ; x\in(\log_25-2;\log_23) " align="absmiddle" class="latex-formula">
Поскольку этот интервал принадлежит ОДЗ, получаем
Ответ: 