log3(7-4x)больше или равно 3 Решите неравенство
ОДЗ 7-4х>0 4x<7<br>x<7/4<br>x<1.75<br> Ответ х∈(-∞; -5]
![log_3(7-4x) \geq 3 \\ \\ log_3(7-4x) \geq Log_33^3 \\ \\ 7-4x \geq 3^3 \\ \\ 7-4x\geq 27 \\ \\ 4x \leq 7-27 \\ \\ 4x \leq -20 \\ \\ x \leq -20:4 \\ \\ x \leq -5 \\ \\ x \in (-\infty; -5]](https://tex.z-dn.net/?f=log_3%287-4x%29+%5Cgeq+3+%5C%5C++%5C%5C+log_3%287-4x%29+%5Cgeq+Log_33%5E3+%5C%5C++%5C%5C+7-4x+%5Cgeq+3%5E3+%5C%5C++%5C%5C+7-4x%5Cgeq+27+%5C%5C++%5C%5C+4x+%5Cleq+7-27+%5C%5C++%5C%5C+4x+%5Cleq+-20+%5C%5C++%5C%5C+x+%5Cleq+-20%3A4+%5C%5C++%5C%5C+x+%5Cleq+-5+%5C%5C++%5C%5C+x+%5Cin+%28-%5Cinfty%3B+-5%5D "log_3(7-4x) \geq 3 \\ \\ log_3(7-4x) \geq Log_33^3 \\ \\ 7-4x \geq 3^3 \\ \\ 7-4x\geq 27 \\ \\ 4x \leq 7-27 \\ \\ 4x \leq -20 \\ \\ x \leq -20:4 \\ \\ x \leq -5 \\ \\ x \in (-\infty; -5]")