Question

Log1/2(x-2)> -4 log1/3(2x-2)> -3 log1/4(3-2x)> -1

Answer 1

$log_{ \frac{1}{2} } (x-2)\ \textgreater \ -4, -4= log_{ \frac{1}{2} } ( \frac{1}{2} ) ^{-4} = log_{ \frac{1}{2} }16$
$log_{ \frac{1}{2} } (x-2)\ \textgreater \ log_{ \frac{1}{2} } 16$
основание логарифма а=1/2, 0<1/2<1
знак неравенства меняем:
$\left \{ {{x-2\ \textless \ 16} \atop {x-2\ \textgreater \ 0}} \right. , \left \{ {{x\ \textless \ 18} \atop {x\ \textgreater \ 2}} \right.$
x∈(2;18)

$log_{ \frac{1}{3} } (2x-2)\ \textgreater \ -3, -3= log_{ \frac{1}{3} } ( \frac{1}{3} ) ^{-3} = log_{ \frac{1}{3} } 27$
$log_{ \frac{1}{3} } (2x-2)\ \textgreater \ log_{ \frac{1}{3} } 27, 0\ \textless \ \frac{1}{3} \ \textless \ 1$
знак неравенства меняем:
$\left \{ {{2x-2\ \textless \ 27} \atop {2x-2\ \textgreater \ 0}} \right. , \left \{ {{2x\ \textless \ 29} \atop {2x\ \textgreater \ 2}} \right. , \left \{ {{x\ \textless \ 14,5} \atop {x\ \textgreater \ 1}} \right.$
x∈(1;14,5)

$log_{ \frac{1}{4} } (3-2x)\ \textgreater \ -1, -1= log_{ \frac{1}{4} } ( \frac{1}{4} ) ^{-1} = log_{ \frac{1}{4} }4$
$log_{ \frac{1}{4} } (3-2x)\ \textgreater \ log_{ \frac{1}{4} } 4, 0\ \textless \ \frac{1}{4}\ \textless \ 1$
знак неравенства меняем:
$\left \{ {{3-2x\ \textless \ 4} \atop {3-2x\ \textgreater \ 0}} \right. , \left \{ {{-2x\ \textless \ 1} \atop {-2x\ \textgreater \ -3}} , \left \{ {{x\ \textgreater \ -0,5} \atop {x\ \textless \ 1,5}} \right. \right.$
x∈(-0,5;1,5)