Question

)решение логарифмических неравенств: 1) 2) 3) 4) 5) 6)

Answer 1

$1) \log_{\frac{1}{3}}(x+4) \ \textgreater \ \log_{\frac{1}{3}}(x^2+2x-2) \\ x+4 \ \textless \ x^2+2x-2\\ x^2+2x-2-x-4\ \textgreater \ 0\\ x^2+x-6\ \textgreater \ 0\\ d=1+24=25\\ x_1=\frac{-1+5}{2}=2; x_2=\frac{-1-5}{2}=-3;$
ответ: $x \in (2,+\infty)$ (т.к. отрицательный корень противоречит определению логарифма)
$2) \lg(2x-3)\ \textgreater \ \lg(x+1)\\ 2x-3\ \textgreater \ x+1\\ 2x-x\ \textgreater \ 1+3\\ x\ \textgreater \ 4\\ \\ 3)\\ \left \{ {{x-18 \ \textless \ 0} \atop {\log_5x \ \textgreater \ 1}} \right. \\ \left \{ {{x \ \textless \ 18} \atop {\log_5x \ \textgreater \ \log_51}} \right. \\ \left \{ {{x\ \textless \ 18} \atop {x \ \textgreater \ 1}} \right. \\ x\in(1,18)\\ \\ 4)\\ \left \{ {{\log_{\frac{1}{3}}x \ \textless \ -2 \atop {x+1\ \textgreater \ 3}} \right.\\$
$\left \{ {{\log_{\frac{1}{3}}x \ \textless \\ \log_{\frac{1}{3}}9 \atop {x\ \textgreater \ 3-1}} \right.\\ \left \{ {{x \ \textless \\ 9 \atop {x\ \textgreater \ 2}} \right.\\ x\in(2,9)\\ 5)\\ \left \{ {{ \ln x \geq =0} \atop {5-x\ \textless \ 0}} \right.\\ \left \{ {{ \ln x \geq =\ln 1} \atop {-x\ \textless \ -5}} \right.\\ \left \{ {{x \geq 1} \atop {x\ \textgreater \ 5}} \right. \\ x\in(5,+\infty)\\ \\$

$6)\\ \left \{ {{\ln x \leq 1} \atop {x+7 \ \textgreater \ 0}} \right. \\ \left \{ {{\ln x \leq \ln 0} \atop {x \ \textgreater \ -7}} \right. \\ \left \{ {{ x \leq 0} \atop {x \ \textgreater \ -7}} \right. \\ x=0$
Т.к. другие значения противоречат определению логарифма