Question

log in baza 5 din 2x-1=2

Answer 1

$\displaystyle log_5(2x-1)=2 \\ 2x-1\ \textgreater \ 0 \Rightarrow x\ \textgreater \ \frac{1}{2} \Rightarrow x \in \left( \frac{1}{2} , \infty \right) \\ log_5(2x-1)=log_55^2 \\ log_5(2x-1)=log_525 \\ 2x-1=25 \\ 2x=25+1 \\ 2x=26 \Rightarrow x= \frac{26}{2} \Rightarrow \boxed{x=13}$