Question

log in baza 3 din (5x-1)=2

Answer 1

Conditia de existenta: 5x-1>0⇒x>1/5⇒x∈(1/5;+∞)
[tex]log_3(5x-1)=2\\ log_3(5x-1)=log_33^2\\ 5x-1=9\\ 5x=10\\ x=2\in( \frac{1}{5} ,+\infty)\\ S=\{2\}[/tex]

Answer 2

$\displaystyle log_3(5x-1)=2~~~~~~~~~~~~~~~~~~~~~~~~ C.E.~5x-1\ \textgreater \ 0 \Rightarrow x\ \textgreater \ \frac{1}{5} \\ \\ log_3(5x-1)=log_39 \\ \\ 5x-1=9 \\ \\ 5x=9+1 \\ \\ 5x=10 \Rightarrow x= \frac{10}{5} \Rightarrow \boxed{x=2}$