Question

Всем . с решением логарифмов а)log0,6(2x-9)> log0.6x б)log3^2(3x)=4+log3(x^4)

Answer 1

$log_{0.6} (2x-9)\ \textgreater \ log_{0.6} x$

ОДЗ:
$\left \{ {{2x-9\ \textgreater \ 0} \atop {x\ \textgreater \ 0}} \right.$

$\left \{ {{2x\ \textgreater \ 9} \atop {x\ \textgreater \ 0}} \right.$

$\left \{ {{x\ \textgreater \ 4.5} \atop {x\ \textgreater \ 0}} \right.$

x∈ $(4.5;+$∞$)$

$2x-9\ \textless \ x$

$2x-x\ \textless \ 9$

$x\ \textless \ 9$

с учетом ОДЗ

ответ: $(4,5;9)$

№2
$log_{3} ^2(3x)=4+log_{3} (x^4)$

ОДЗ:
$\left \{ {{3x\ \textgreater \ 0} \atop {x^4\ \textgreater \ 0}} \right.$

$\left \{ {{x\ \textgreater \ 0} \atop {x \neq 0}} \right.$

x∈ $(0;+$∞$)$

$( log_{3} 3+ log_{3} x)^2=4+4 log_{3}x$

$( 1+ log_{3} x)^2=4+4 log_{3}x$

$1+ log_{3}^2 x+2 log_{3} x-4-4 log_{3} x=0$

$log_{3}^2 x-2 log_{3} x-3=0$

Замена: $log_{3} x=t$

$t^2-2t-3=0$

$D=4+12=16$

$t_1=3$
$t_2=-1$

$log_{3} x=3$  или    $log_{3} x=-1$

$x=27$   или $x= \frac{1}{3}$