Решить неравенство. логарифм корня из 3х+4 по основанию 2 умножить на логарифм 2 по основанию х больше единицы.

Ответы

ник200781 17.09.2020 23:51

$log_2\sqrt{3x+4}\cdot log_{x}2\ \textgreater \ 1\; ,\; \; OOF:\; \; \left \{ {{3x+4\ \textgreater \ 0} \atop {x\ \textgreater \ 0,\; x\ne 1}} \right. \\\\\frac{log_2\sqrt{3x+4}}{log_2x}\ \textgreater \ 1\; \; \Rightarrow \; \\\\a) \left \{ {{log_2x\ \textgreater \ 0} \atop {log_2\sqrt{3x+4}\ \textgreater \ log_2x}} \right. \; ,\; \left \{ {{x\ \textgreater \ 1} \atop {\sqrt{3x+4}\ \textgreater \ x}} \right. \\\\\sqrt{3x+4}\ \textgreater \ x\; \to \; \left \{ {{3x+4\ \textgreater \ x^2} \atop {x 0}} \right. \; \; ili\; \; \left \{ {{3x+4 \geq 0} \atop {x\ \textless \ 0\; }} \right. \; \; (x\ \textless \ 0\; \notin OOF)$

$\left \{ {{x^2-3x-4\ \textless \ 0} \atop {x 0}} \right. \;$ $\left \{ {{-1\ \textless \ x\ \textless \ 4} \atop {x 0,x\ne 1}} \right. \; \to \; x\in ( 0,1)\cup (1,4)$

$b) \left \{ {{log_2x\ \textless \ 0} \atop {log_2\sqrt{3x+4}\ \textless \ log_2x}} \right. \; ,\; \left \{ {{0\ \textless \ x\ \textless \ 1} \atop {\sqrt{3x+4}\ \textless \ x}} \right. \; ,\; \left \{ {{0\ \textless \ x\ \textless \ 1} \atop {3x+4\ \textless \ x^2,\; 3x+4 \geq 0,\; x\ \textgreater \ 0}} \right. \; ,\\\\x^2-3x-4\ \textgreater \ 0\; \; \to \; \; x\in (-\infty ,-1)\cup (4,+\infty )\; ,\; \; x\ \textgreater \ 0\\\\ \left \{ {{0\ \textless \ x\ \textless \ 1} \atop {x\in (4,+\infty )}} \right. \; \; \to \; \; x\in \varnothing$

ответ: $x\in( 0,1)\cup (1,4)$ .