Question

Log2(2x+3)>log(x-2),
Log6(3x-1) log6(9x+4);

Answer 1

$\begin {cases} log_2\left(2x+3\right)>\:log_2\left(x-2\right) \\ log_6\left(3x-1\right)\le \:log_6\left(9x+4\right) \end.\\\\ \\log_2\left(2x+3\right)>\:log_2\left(x-2\right)\\\\\begin {cases} 2x+3>x-2 \\ x-2>0 \end.\\\\\begin {cases} x>-5 \\ \:x>2 \end.\\\\x>2\\\\log_6\left(3x-1\right)\le \:log_6\left(9x+4\right)\\\\\begin {cases} 3x-1\le \:9x+4 \\ \:3x-1>0 \end.\\\\\begin {cases} -6x\le \:5 \\ \:3x>1 \end.\\\\\begin {cases} x\ge \:-\frac{5}{6} \\ \:x>\frac{1}{3} \end.\\\\x>\frac{1}{3}$

$\begin {cases} x>2 \\ x>\frac{1}{3} \end.\\x>2$

Ответ: x > 2.