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Determinati toate functiile continue \( f:{\mathbb R} \rightarrow (0, \infty) \) cu proprietatea \( f(2x-1)\cdot f(3x+1)=e^{6x-5}, \forall x \in {\mathbb R} \).

Fie functia \( g:\mathbb{R}\to (0,\infty) \),\( g(x)=\ln f(x)-\frac{6}{5}x+\frac{5}{2} \).
Logaritmand si trecand tot in partea stanga relatia din ipoteza devine: \( g(2x-1)+g(3x+1)=0 \).
Daca in locul lui x introducem acum valoarea \( \frac{x-1}{3} \) avem:

\( g(\frac{2}{3}x-\frac{5}{2})+g(x)=0 \) de unde rezulta \( g(x)=-g(\frac{2}{3}x-\frac{5}{2}) \). Prin inductie rezulta:

\( g(x)=(-1)^{n-1}g\left(\left(\frac{2}{3}\right)^{n-1}x -\frac{15}{2}\left(1-\left(\frac{2}{3}\right)^{n-1}\right)\right) ,\forall n\in\mathbb{N} ,n\geq 2 \). Pentru n tinzand la infinit rezulta \( g(x)=c \) unde \( c\in \{ -g(-\frac{15}{2}),\ g(-\frac{15}{2}) \} \forall x \in \mathbb{R} \). Dar \( g(-2)=0 \) de unde avem c=0, deci \( g(x)=0 \forall x\in\mathbb{R} \). Asadar \( f(x)=e^{\frac{6}{5}x-\frac{5}{2}} \).