Question

решите:
$log _{2x} 0.25 leq log_{2} 32x-1 log_{2}(2^{x}-1)*log _{1/2}(2^{x+1} - 2) textgreater -2$

Answer 1

$1)quad log_{2x}0,25 leq log_232x-1; ;; ; ODZ:; ; x textgreater 0,; xne 1\\star ; ; log_{2x}0,25=log_{2x}frac{1}{4}=log_{2x}2^{-2}=-2log_{2x}2=frac{-2}{log_22x}=\\=frac{-2}{log_22+log_2x}=frac{-2}{1+log_2x};\\star ; ; log_232x=log_2(2^5x)=log_22^5+log_2x=5+log_2x\\\t=log_2x; ,; ; frac{-2}{1+t }leq 5+t-1; ,\\frac{-2}{1+t}-t-4 leq 0; ,; ; frac{-2-(t+4)(1+t)}{1+t} leq 0; ,\\ frac{-2-(t+t^2+4+4t)}{t+1} leq 0; ,; ; frac{-(t^2+5t+6)}{t+1} leq 0; ,$

$frac{t^2+5t+6}{t+1 }geq 0 ; ,; ; frac{(t+2)(t+3)}{t+1} geq 0\\Znaki; drobi:; ; ---[-3, ]+++[-2, ]---(-1)+++\\tin [-3,-2, ]cup (-1,+infty )qquad Rightarrow \\t=log_2x:; ; left [ {{-3 leq log_2x leq -2} atop {log_2x textgreater -1}} right. ; left [ {{log_22^{-3} leq log_2x leq log_22^{-2}} atop {log_2x textgreater log_22^{-1}}} right. ; left [ {{frac{1}{8} leq x leq frac{1}{4}} atop {x textgreater frac{1}{2}}} right. \\xin [, frac{1}{8},frac{1}{4}, ]cup (frac{1}{2},1)cup(1,+infty )$

$2)quad log_2(2^{x}-1)cdot log_{frac{1}{2}}(2^{x+1}-2) textgreater -2; ,; ; ODZ:; xin R\\star ; ; log_{frac{1}{2}}(2^{x+1}-2)=-log_2(2cdot (2^{x}-1))=-(log_22+log_2(2^{x}-1))=\\=-1-log_2(2^{x}-1); ; star \\\t=log_2(2^{x}-1); ; to ; ; ; ; tcdot (-1-t) textgreater -2; ,\\-t-t^2+2 textgreater 0; ,; ; ; t^2+t-2 textless 0; ,\\D=1+4cdot 2=9; ,; ; t_1=frac{-1-3}{2}=-2; ,; ; t_2=frac{-1+3}{2}=1\\a); ; log_2(2^{x}-1)=-2; to ; ; 2^{x}-1=2^{-2}; ,; ; 2^{x}=1+frac{1}{4}; ,; 2^{x}=frac{5}{4}$

$x=log_2frac{5}{4}; ,; ; x=log_25-log_24=log_25-log_22^2=log_25-2\\b); ; log_2(2^{x}-1)=1; to ; ; ; 2^{x}-1=2^1; ,; ; 2^{x}=3\\x=log_23\\Otvet:; ; x=log_25-2; ,; ; x=log_23; .$