Question

Log5 (x^2 -3x -4) + 2/(log3(0,2) < 0

Answer 1

$log_5 (x^2-3x-4)+\frac{2}{log_3 0.2} \ \textless \ 0 \\ x^2-3x-4\ \textgreater \ 0; (x-4)(x+1)\ \textgreater \ 0; \left \{ {{x\ \textgreater \ 4} \atop {x\ \textless \ -1}} \right. \\ x\in (-\infty;-1) \cup (4;+\infty) \\\\ log_5 (x^2-3x-4)+\frac{2}{log_3 \frac{1}{5} } \ \textless \ 0 \\ log_5 (x^2-3x-4)-\frac{2}{log_3 5 } \ \textless \ 0 \\ log_5 (x^2-3x-4)-2log_5 3\ \textless \ 0 \\ log_5 \frac{x^2-3x-4}{9} \ \textless \ log_5 1 \\ \frac{x^2-3x-4}{9} \ \textless \ 1 \\ x^2-3x-13 \ \textless \ 0 \\ D=(-3)^2-4*(-13)=9+52=61 \\ x=\frac{3\pm\sqrt{61}}{2} \\ (x-\frac{3+\sqrt{61}}{2})(x-\frac{3-\sqrt{61}}{2})\ \textless \ 0$
$\\ \frac{3-\sqrt{61}}{2}\ \textless \ x\ \textless \ \frac{3+\sqrt{61}}{2} \\\\ x\in (\frac{3-\sqrt{61}}{2};-1)\cup(4;\frac{3+\sqrt{61}}{2})$