Срочно, решите пожалуйсто!

Поделись с друзьями

1. $log_{26}2+log_{26}13=log_{26}(2*13)=log_{26}26=1$
2. а) log_4(2x+10)=log_4(6x-2)
$2x+10\ \textgreater \ 0; 6x-2\ \textgreater \ 0$
$x\ \textgreater \ -5;x\ \textgreater \ \frac{1}{3}$
2x+10=6x-2
4x=12
x=3
б) log_8(5x-1)=2
$5x-1\ \textgreater \ 0$
$x\ \textgreater \ \frac{1}{5}$
5x-1=8^2
5x=65
x=13
в) log_5(2x+1)+log_5(16x-7)=3
$2x+1\ \textgreater \ 0;16x-7\ \textgreater \ 0$
$x\ \textgreater \ - \frac{1}{2} ;x\ \textgreater \ \frac{7}{16}$
log_5 ((2x+1)*(16x-7))=log_55^3
(2x+1)(16x-7)=125
$32 x^{2} -14x+16x-7-125=0$
$32 x^{2} +2x-132=0$
$16 x^{2} +x-66=0$
D=1-4*16*(-66)=1+4224=4225
$\sqrt{D}=65$
$x_1= \frac{-1+65}{2*16} = \frac{64}{32}=2;x_2= \frac{-1-65}{2*16} = \frac{-66}{32}=-2 \frac{1}{16}$
x_2 не входит в ОДЗ.
г) log_3(5x+1)-log_3(x+1)=1
$5x+1\ \textgreater \ 0;x+1\ \textgreater \ 0$
$x\ \textgreater \ - \frac{1}{5} ; x\ \textgreater \ -1$
$log_3 \frac{5x+1}{x+1}=log_33$
5x+1=3x+3
2x=2
x=1