Question

100 баллов! срочно
решить неравенство
$log_{x}(3x) \leqslant \sqrt{ log_{x}(3 {x}^{7} ) }$
​

Answer 1

Ответ:

$x \in \left( {0;\,\,\displaystyle\frac{1}{{\sqrt[7]{3}}}} \right]\cup \left[ {\sqrt 3 ;\,\, + \infty } \right)$

Объяснение:

Пользуясь свойствами логарифма

${\log _a}a = 1,$

${\log _a}{b^n} = n{\log _a}b\ (b > 0),$

${\log _a}(xy) = {\log _a}x + {\log _a}y\ (x,\,\,y > 0),$

получим:

${\log _x}3 + {\log _x}x \le \sqrt {{{\log }_x}3 + {{\log }_x}{x^7}} ;\\\\{\log _x}3 + 1 \le \sqrt {{{\log }_x}3 + 7} .$

Сделаем замену ${\log _x}3 + 1 = t,$ тогда

$t \le \sqrt {t + 6} ;\\\\\left[ \begin{array}{l}\left\{ \begin{array}{l}t < 0,\\t + 6 \ge 0,\end{array} \right.\\\left\{ \begin{array}{l}t \ge 0,\\{t^2} \le t + 6;\end{array} \right.\end{array} \right.$

$\left[ \begin{array}{l}\left\{ \begin{array}{l}t < 0,\\t \ge - 6,\end{array} \right.\\\left\{ \begin{array}{l}t \ge 0,\\{t^2} - t - 6 \le 0;\end{array} \right.\end{array} \right.\\\\\left[ \begin{array}{l}t \in [ - 6;\,\,0),\\t \in [0;\,\,3];\end{array} \right.\\\\t \in [ - 6;\,\,3].$

Делая обратную замену, получим:

$\left\{ \begin{array}{l}{\log _x}3 + 1 \ge - 6,\\{\log _x}3 + 1 \le 3;\end{array} \right.\\\\\left\{ \begin{array}{l}{\log _x}3 \ge - 7,\\{\log _x}3 \le 2;\end{array} \right.\\\\\left\{ \begin{array}{l}{\log _x}3 \ge {\log _x}{x^{ - 7}},\\{\log _x}3 \le {\log _x}{x^2};\end{array} \right.$

$\left[ \begin{array}{l}\left\{ \begin{array}{l}0 < x < 1,\\{x^{ - 7}} \ge 3,\\{x^2} \le 3,\end{array} \right.\\\left\{ \begin{array}{l}x > 1,\\{x^{ - 7}} \le 3,\\{x^2} \ge 3;\end{array} \right.\end{array} \right.$

$\left\{ \begin{array}{l}0 < x < 1,\\x \le \displaystyle\frac{1}{{\sqrt[7]{3}}},\\x \in [ - \sqrt 3 ;\,\,\sqrt 3 ],\end{array} \right.\\\\\left\{ \begin{array}{l}x > 1,\\x \ge \displaystyle\frac{1}{{\sqrt[7]{3}}},\\x \in \left( { - \infty ;\,\, - \sqrt 3 } \right] \cup \left[ {\sqrt 3 ;\,\, + \infty } \right);\end{array} \right.\\\\\left[ \begin{array}{l}x \in \left( {0;\,\,\displaystyle\frac{1}{{\sqrt[7]{3}}}} \right],\\x \in \left[ {\sqrt 3 ;\,\, + \infty } \right).\end{array} \right.$