Conditii de existenta
x-1>0=>x>1=> D = (1,+ infinit)
3lg(x-1)-6=lg5-lg(x-1)
duci logaritmii in dreapta si 6 in stanga
3lg(x-1)+lg(x-1)-lg5=6
4lg(x-1)-lg5=6
aplici proprietatea scaderii logaritmilor
4(x-1) impartit la 5 = 6
(4x-4)/5=6=>4x-4=6=>4x=10=>x=10/4=>x=5/2 apartine D S={5/2}
