Question

log 3 ( x+2) - log 1/3 ( x-6) = log 3 ( 2x-5)?

Answer 1

$\log_3(x+2)-\log_{1/3}(x-6)=\log_{3}(2x-5)\\ \\ \log_3(x+2)+\log_{(1/3)^{-1}}(x-6)=\log_{3}(2x-5)\\ \\ \log_3(x+2)+\log_{3}(x-6)=\log_{3}(2x-5)\\ \\ \log_3[(x+2)(x-6)]=\log_{3}(2x-5)\\ \\ (x+2)(x-6)=2x-5\\ \\ x^2-4x-12=2x-5\\ \\ x^2-6x-7=0\\ \\ (x-7)(x+1)=0\\ \\ \\ x\in\{-1,7\}$

Veamos los argumentos de cada logaritmo;

$x+2\ \textgreater \ 0\to x\ \textgreater \ -2\\ \\ x-6\ \textgreater \ 0\to x\ \textgreater \ 6\\ \\ 2x-5\ \textgreater \ 0\to x\ \textgreater \ \dfrac{5}{2}\\ \\ \text{En resumen: }\mathbf{x\ \textgreater \ 6}$

Por lo tanto $\boxed{\boxed{\mathbf{x=7}}}$