Question

9^(log3(x))^2<4x^log3(x)-3

Answer 1

$9^{log_3^2x}<4\cdot x^{log_3x}-3\; \; ,\; \; ODZ:\; \; x>0\; ,\; x\ne 1\\\\9^{log_3^2x}=(3^2)^{log_3x\cdot log_3x}=3^{2\cdot log_3x\cdot log_3x}=3^{log_3{x^2}\cdot log_3x}=\\\\=(3^{log_3x^2})^{log_3x}=(x^2)^{log_3x}=x^{2log_3x}=(x^{log_3x})^2\\\\t=x^{log_3x}>0\; ,\; \; t^2<4t-3\; \; \to \; \; t^2-4t+3<0\; ,\\\\ t_1=1\; ,\; t_2=3\; \; ,\; \; (t-1)(t-3)<0\\\\znaki:\; \; \; +++(1)---(3)+++\quad t\in (1,3)\\\\\left \{ {{x^{log_3x}>1} \atop {x^{log_3x}<3}} \right. \; \left \{ {{log_3(x^{log_3x})>log_31} \atop {log_3(x^{log_3x})<log_33}} \right. \; \left \{ {{log_3^2x>0} \atop {log_3^2x<1}} \right. \; \left \{ {{log_3x\ne 0 } \atop {-1<log_3x<1}} \right. \;$

$\left \{ {{x\ne 1} \atop {\frac{1}{3}<x<3}} \right. \; \; \Rightarrow \; \; x\in (\frac{1}{3},1)\cup (1,3)$