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)