Question

2log_3(x+1)-2=log_3(x-1) решить !

Answer 1

ОДЗ:
$\left \{ {{x+1\ \textgreater \ 0} \atop {x-1\ \textgreater \ 0}} \right.$

$\left \{ {{x\ \textgreater \ -1} \atop {x\ \textgreater \ 1}} \right.$

$x$ ∈ $(1;+$ ∞ $)$

$2log_3(x+1)=log_3(x-1)+2$

$log_3(x+1)^2=log_3(x-1)+log_39$

$log_3(x+1)^2=log_3[9*(x-1)]$

$(x+1)^2=9*(x-1)$

$x^2+2x+1=9x-9$

$x^2+2x+1-9x+9=0$

$x^2-7x+10=0$

$D=(-7)^2-4*1*10=49-40=9$

$x_1= \frac{7+3}{2}=5$

$x_2= \frac{7-3}{2}=2$

ответ: $2;$  $5$