Question

логарифмическое неравенство. я решила, но меня смущает ответ. хочу свериться

Answer 1

$[log_{frac{1}{3}}(-log_2(x))]^2-log_{frac{1}{3}}[(log_2(x))^2] geq 15$

$left { {{[log_{{3}^{-1}}(-t)]^2-log_{{3}^{-1}}(t^2) geq 15} atop {t=log_2(x)}} right.\ \ left { {{[-log_{3}(-t)]^2+log_{3}((-t)^2) geq 15} atop {t=log_2(x)}} right. \ \ left { {{[log_{3}(z)]^2+log_{3}(z^2) geq 15} atop {z=-t=-log_2(x)}} right. \ \ left { {{[log_{3}(z)]^2+2log_{3}(z) - 15 geq 0} atop {z=-log_2(x)}} right. \ \ left { {{k^2+2k - 15 geq 0} atop {k=log_3(z) and z=-log_2(x)}} right.$

$left { {{[k-(-5)]*[k-3] geq 0} atop {k=log_3(z) and z=-log_2(x)}} right. \ \ left { {{k leq -5 or k geq 3} atop {k=log_3(z) and z=-log_2(x)}} right. \\ left { {{log_3(z) leq log_3(3^{-5}) or log_3(z) geq log_3(3^3)} atop {z=-log_2(x)}} right. \ \ left { {{(z leq 3^{-5} or z geq 3^3) and z>0} atop {z=-log_2(x)}} right.$

$(-log_2(x) leq 3^{-5} or -log_2(x) geq 3^3) and -log_2(x) textgreater 0\ \ left { {{log_2(x) geq -3^{-5}*log_2(2) or log_2(x)) leq -3^3*log_2(2)} atop {log_2(x) textless 0}} right.$

$left { {{x geq 2^{-3^{-5}} or x leq 2^{-3^3}} atop {log_2(x) textless log_2(1)}} right.\ \ left { {{x geq 2^{- frac{1}{243} } or x leq 2^{-27}} atop {x textless 1 and x textgreater 0}} right.$

$2^{- frac{1}{243} }>2^{-27}$
$2^{- frac{1}{243} } textless 1$

$left { {{x geq 2^{- frac{1}{243} } or x leq 2^{-27}} atop {x textless 1 and x textgreater 0}} right. \ \ left { {{xin (-infty; 2^{-27}] cup[2^{- frac{1}{243}}; +infty })} atop {xin (0; 1)}} right. \ \ xin (0; 2^{-27}] cup [2^{- frac{1}{243}}; 1)$