Question

$Rezolvati~ecuatia: \\ log_{3} {(2x-1)}~*(1+log_{3}{(3-x)} )=log_{(x+1)} 3$

Answer 1

$\displaystyle Punand~conditiile~de~existenta,~obtinem~x \in \left(\frac{1}{2},3 \right). \\ \\ Rescriem~ecuatia~in~felul~urmator: \\ \\ \log_3(x+1) \cdot \log_3(2x-1) \cdot \log_3(9-3x)=1.~(*) \\ \\ Observam~ca~daca~x \in \left( \frac{1}{2},1 \right],~atunci~avem~\log_3(x+1)\ \textgreater \ 0, \\ \\ \log_3(2x-1) \le 0~si~\log_3(9-3x)\ \textgreater \ 0,~deci \\ \\ \log_3(x+1) \cdot \log_3(2x-1) \cdot \log_3(9-3x) \le 0. \\ \\ Rezulta,~deci~ca~x \in(1,3),~caz~in~care~fiecare~logaritm~din ~(*) \\ \\ este~pozitiv.$

$\displaystyle Din~inegalitatea~mediilor,~avem: \\ \\ \log_3(x+1) \cdot \log_3(2x-1) \cdot \log_3(9-3x) \le \\ \\ \left( \frac{ \log_3(x+1) + \log_3(2x-1) + \log_3(9-3x)}{3}\right)^3=E. \\ \\ Din~inegalitatea~lui~Jensen~aplicata~functiei~concave \\ \\ f: (0,+ \infty) \to \mathbb{R},~f(x)=\log_3x~obtinem \\ \\ E \le \log_3^3 \frac{(x+1)+(2x-1)+(9-3x)}{3}=log_3^33=1.$

$\displaystyle Si~pentru~ca~trebuie~sa~avem~egalitate,~este~necesar~ca \\ \\ \log_3(x+1)=\log_3(2x-1)=\log_3(9-3x). \\ \\ Adica~x+1=2x-1=9-3x. \\ \\ Adica~x=2.$