Question

Допоможіть розвязати нерівності*)) що зможите))
1) log 3 (2x+1) < log 3 ( x-1)
2)log 1/3 (2-x ) > 0
3) log 31 ( 31x+2) <1
4) log 1/11 ( 2x-1) + log1/11 x>0

Answer 1

$1) log_3(2x+1) textless log_3(x-1)\3 textgreater 1 = textgreater 2x+1 textless x-1\x textless -2\D: left { {{2x+1 textgreater 0} atop {x-1 textgreater 0}} right. ,left { {{x textgreater -frac{1}{2}} atop {x textgreater 1}} right. , x textgreater 1\ \ left { {{x textless -2} atop {x textgreater 1}} right.$
нет решений

$2) log_{frac{1}{3}}(2-x) textgreater 0\log_{frac{1}{3}}(2-x) textgreater log_{frac{1}{3}}1\frac{1}{3} textless 1 = textgreater 2-x textless 1\x textgreater 1\ \D: 2-x textgreater 0, x textless 2\ \ left { {{x textgreater 1} atop {x textless 2}} right. 1 textless x textless 2$

$3) log_{31}(31x+2) textless 1\log_{31}(31x+2) textless log_{31}31\31 textgreater 1 = textgreater 31x+2 textgreater 31\31x textgreater 29\x textgreater frac{29}{31}\ \ D: 31x+2 textgreater 0, 31x textgreater -2, x textgreater -frac{2}{31}\ \ left { {{x textgreater frac{29}{31}} atop {x textgreater -frac{2}{31}}} right. x textgreater frac{29}{31}$

$4) log_{frac{1}{11}}(2x-1)+log_{frac{1}{11}}x textgreater 0\log_{frac{1}{11}}(2x-1) textgreater -log_{frac{1}{11}}x\log_{frac{1}{11}}(2x-1) textgreater log_{frac{1}{11}}x^{-1}\log_{frac{1}{11}}(2x-1) textgreater log_{frac{1}{11}}frac{1}{x}\frac{1}{11} textless 1= textgreater 2x-1 textless frac{1}{x}\frac{2x^2-x-1}{x} textless 0\frac{2(x-1)(x+frac{1}{2})}{x} textless 0\x textless -frac{1}{2}, x textgreater 1$
$D: left { {{2x-1 textgreater 0} atop {x textgreater 0}} right. \ left { {{x textgreater frac{1}{2}} atop {x textgreater 0}} right. ,x textgreater frac{1}{2}\ \ left { {{x textless -frac{1}{2}, x textgreater 1} atop {x textgreater frac{1}{2}}} right. , x textgreater 1$