conditii existenta: 3x+1>0 si x-1 >0
log₅ (3x+1)=1+log₅ (x-1)
log₅(3x+1)-log₅(x-1)=1
[tex]log_{5} \frac{3x+1}{x-1} =1 \\ log_{5} \frac{3x+1}{x-1} =log_{5} 5 [/tex]
se elimina bazele,fiindca baza e 5 apartine multimii N
[tex] \frac{3x+1}{x-1}=5 =\ \textgreater \ 3x+1=5(x-1) =\ \textgreater \ 3x+1=5x-5 =\ \textgreater \ -2x=-6 \\ x=3 [/tex]
verific conditiile: 3*3+1=10> 0 (adevarat)
3-1=2 > 0(adevarat)
solutie finala: x=3
