Question

Urgenta !!!

Log din 5 (3x+1)=1+log don 5 (x-1)

Answer 1

conditii existenta:  3x+1>0  si x-1 >0
log₅ (3x+1)=1+log₅ (x-1)
log₅(3x+1)-log₅(x-1)=1
$log_{5} \frac{3x+1}{x-1} =1 \\ log_{5} \frac{3x+1}{x-1} =log_{5} 5$

se elimina bazele,fiindca baza e 5 apartine multimii N

$\frac{3x+1}{x-1}=5 =\ \textgreater \ 3x+1=5(x-1) =\ \textgreater \ 3x+1=5x-5 =\ \textgreater \ -2x=-6 \\ x=3$

verific conditiile:  3*3+1=10> 0 (adevarat)
                          3-1=2  > 0(adevarat)

solutie finala: x=3