Question

1) Log3(2x-4) больше Log3(14-x)

2)Log1/3(x+15)≥Log1/3(x-1)-2

Answer 1

1.

$log_{3}(2x - 4) > log_{3}(14 - x)$

ОДЗ:

$\left \{ {{2x - 4 > 0} \atop {14 - x > 0} } \right. \\ \\ \left \{ {{x > 2} \atop {x < 14} } \right. \\ \\ = > x\in(2;14)$

$log_{3}(2x - 4) > log_{3}(14 - x) \\ 2x - 4 > 14 - x \\ 3x > 18 \\ x > 6$

С ОДЗ:

$x\in(6; 14)$

2.

$log_{ \frac{1}{3} }(x + 15) \geqslant log_{ \frac{1}{3} }( x-1) - 2$

ОДЗ:

$\left \{ {{x + 15 > 0} \atop {x-1 > 0} } \right. \\ \\ \left \{ {{ x > - 15} \atop {x > 1} } \right. \\ \\ =>x>1$

$log_{ \frac{1}{3} }(x + 15) \geqslant log_{ \frac{1}{3} }(x-1) - 2 \\ log_{ \frac{1}{3} }(x + 15) \geqslant log_{ \frac{1}{3} }(x-1) - log_{ \frac{1}{3} }( \frac{1}{9} ) \\ log_{ \frac{1}{3} }(x + 15) \geqslant log_{ \frac{1}{3} }(9(x-1)) \\ \\ \frac{1}{3} < 1 \\ \text{знак меняется} \\ \\ x + 15 \leqslant 9(x-1) \\ x + 15\leqslant 9x -9\\ -8x \leqslant -24 \\ x \geqslant 3$

С ОДЗ:

$x \in[3;+\infty)$