Question

como resolver, log5x=logx5?

Answer 1

$Ol\acute{a}~~\mathbb{VITORIA} \\ \\ Seja: \\ \\ log_5x=log_x5 \\ \\ log_5x-log_x5=0~~---\ \textgreater \ por~propiedade~[log_ab= \frac{log_mb}{log_ma}], temos. \\ \\ log_5x- \frac{log_55}{log_5x}=0~~--\ \textgreater \ [log_55=1] \\ \\ Log_5x- \frac{1}{log_5x} =0 \\ \\ log_5x.(log_5x)-1=0$


$(log_5x)\²=1 \\ \\ Log_5x=\pm~ \sqrt{1} \\ \\ log_5x=\pm~1~~--\ \textgreater \ por~propriedade~[log_ab=n=\ \textgreater \ a^{n}=b] \\ \\ \boxed{log_5x=\pm1}~\begin{cases}log_5x=+1=\ \textgreater \ 5^{1} =x \\ ~~~~~~~~~~~~~~~~~~~~~\boxed{x'=5}\\ \\ log_5x=-1=\ \textgreater \ 5^{-1}=x \\ ~~~~~~~~~~~~~~~~~~~~~~\boxed{x''= \frac{1}{5}} \end{cases} \\ \\ Temos~2~valores~para~(x)\\ \therefore~C.S~\{5, \frac{1}{5}~\}~$

$\mathbb{vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~ter~ajudado!!$