Question

Решить 3lg^2x-5lgx+2=0 log7(5x-4)< 1 log3x+2log9x+3log27x< или=3

Answer 1

Первый номер. 

$3lg^2x-5lgx+2=0$
$lgx=a$, тогда $3a^2-5a+2=0$
$D=25-24=1\\a_{1,2}=\frac{5б1}{6}\to\left[\begin{array}{ccc}lgx=a_1=1\\lgx=a_2=\frac{2}{3}\end{array}\right\to\left[\begin{array}{ccc}x_1=10\\x_2=\sqrt[3]{100}\end{array}\right$

Второй номер. 

$log_7(5x-4)\ \textless \ 1\to\left[\begin{array}{ccc}5x-4\ \textless \ 7\\5x-4\ \textgreater \ 0\end{array}\right\to\left[\begin{array}{ccc}5x\ \textless \ 11\\5x\ \textgreater \ 4\end{array}\right\to\left[\begin{array}{ccc}x\ \textless \ 2,2\\x\ \textgreater \ 0,8\end{array}\right\\0,8\ \textless \ x\ \textless \ 2,2$
ответ: x∈(0,8; 2,2)

Третий номер. 

$log_3x+2log_9x+3log_{27}x\leq3\\log_3x+log_3x+log_3x\leq3\\3log_3x\leq3\\log_3x\leq1\to\left[\begin{array}{ccc}x\leq3\\x\ \textgreater \ 0\end{array}\right\\0\ \textless \ x\leq3$
ответ: x∈(0; 3]