Question

log2

3

(x − 1) − 2log3

(x − 1) > 3​

Answer 1

$\displaystyle log_3^2(x-1) - 2log_3(x-1)>3$

ОДЗ: x-1 > 0     =>     x>1

$\displaystyle log_3^2(x-1) - 2log_3(x-1)>3\\\bigg| log_3(x-1) = t \bigg|\\\\t^2 - 2t>3\\t^2-2t-3>0\\D = 4-4*(-3*1) = 16\\\sqrt{D} = 4\\\\t=\frac{2\pm4}{2}\\t_1 = 3\\t_2 = -1$

$\displastyle \ _{} ~~~~~~~~~~ +~~~~~~~~-~~~~~~~+\\-\infty \ \ \text{---------o}\text{---------o}\text{---------}>\text{t}\\ \ \ ~~~~~~~~~~~~~~~~~\text{-1}~~~~~~~~~3 \\\\ t\in(-\infty;-1)\cup(3;+\infty)$

$\displaystyle \left[\begin{gathered}log_3(x-1)<-1\\log_3(x-1)> \ \ 3\end{gathered}\right. \hfill _{}\\\\\\\left[\begin{gathered}x-1<\frac{1}{3}\\x-1>27\end{gathered}\right. _{} ->\left[\begin{gathered}x<\frac{4}{3}\\x>28\end{gathered}\right. \hfill \ _{}_{}$

Найдём пересечение с ОДЗ:

$\\~~_{_{\text{------------------------------}}}_{_{\text{------------------------}}}_{_{\text{--------------------}}} \\ ~~\diagdown\diagdown\diagdown\diagdown\diagdown\diagdow\diagdown\big{|}\big\times\big\times\big\times\big\times\big\time\big{|}\diagup\diagup\diagup\diagup\big{|}\big\times\big\times\big\times\big\times\big\times \\-\infty \ \ \text{---------o}\text{---------o}\text{---------o}\text{---------}>+\infty\\ \ \ ~~~~~~~~~~~~~~~~~~\text{1}~~~~~~~~~\frac{4}{3} ~~~~~~~~\text{2}\text{8}$

$\displaystyle x\in\bigg(1;\frac{4}{3}\bigg)\cup\bigg(28;+\infty\bigg)$