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

Question 1

Question 2

в таком неравенстве ответ вряд ли будет "нормальным"... 0 < x <= 1:2^27 и 1:(243V2) <= x < 1

Question 3

в скобочках --это корень 243-й степени их двух или два в степени (1/243)

Question 4

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

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

0} \atop {z=-log_2(x)}} \right." alt=" \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." align="absmiddle" class="latex-formula">

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

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

2^{-27}" alt="2^{- \frac{1}{243} }>2^{-27}" align="absmiddle" class="latex-formula">
$2^{- \frac{1}{243} }\ \textless \ 1$

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

xtoto_zn · Answer 1 · 2018-05-27T11:57:23+0000

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

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

0} \atop {z=-log_2(x)}} \right." alt=" \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." align="absmiddle" class="latex-formula">

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

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

2^{-27}" alt="2^{- \frac{1}{243} }>2^{-27}" align="absmiddle" class="latex-formula">
$2^{- \frac{1}{243} }\ \textless \ 1$

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