Question

vă rog ajutați mă,e urgent

Answer 1

Nu inteleg ce scrie la p

Answer 2

$p) \ log_5(2x-1)=log_53$
$2x-1\ \textgreater \ 0 \ \ \ x\ \textgreater \ \frac{1}{2}$
$fct.logaritmica \ este \ injectiva \rightarrow 2x-1=3$
$2x=4 \rightarrow x=2 \ \ \textgreater \ \frac{1}{2} \Rightarrow S={2}$
$q) log_5(x-3)^2=log_5(x-1)$
$\left \{ {{x-3 \ \textgreater \ 0 \ \ \ x\ \textgreater \ 3} \atop {x-1\ \textgreater \ 0 \ \ \ x\ \textgreater \ 1}} \right. \Rightarrow x \ \textgreater \ 3$
$fct.logaritmica \ este \ injectiva \rightarrow (x-3)^2=x-1$
[tex](x-3)^2=x-1 [/tex]
$x^2-6x+9-x+1=0$
$x^2-7x+10=0$
$\delta=b^2-4ac=(-7)^2-4*1*10= 49-40=9$
$\left \{ {{x_1= \frac{-b+ \sqrt{\delta} }{2a} } \atop {x_1= \frac{-b- \sqrt{\delta} }{2a}}} \right. \rightarrow \left \{ {{x_1= \frac{7+3}{2} } \atop {x_1= \frac{7-3}{2}}} \right. \rightarrow \left \{ {{x_1=5 \ \ \textgreater \ 3 \ (A)} \atop {x_2=2 \ \ \textgreater \ 3 \ (F)}} \right. \Rightarrow S={5}$