Question

log5 (1-x) + log5 (2-x)= log5 (8-2x)
Como posso resolver?

Answer 1

$log_5(1-x)+log_5(2-x)=log_5(8-2x)\\ log_5(1-x).(2-x)=log_5(8-2x)\\ (1-x).(2-x)=(8-2x)\\ 2-x-2x+x^2=8-2x\\ x^2-x-6=0\\ \Delta=(-1)^2-4.1.(-6)=1+24=25\\ x=\frac{1\pm\sqrt{25}}{2}\\\\ x_1=\frac{1+5}{2}=3\\\\ x_2=\frac{1-5}{2}=-2\\\\ 1-x\ \textgreater \ 0\\ 1\ \textgreater \ x\\ x\ \textless \ 1\\\\ 2-x\ \textgreater \ 0\\ 2\ \textgreater \ x\\ x\ \textless \ 2\\\\ 8-2x\ \textgreater \ 0\\ 8\ \textgreater \ 2x\\ 4\ \textgreater \ x\\ x\ \textless \ 4\\\\ Resposta: x=-2$