C.E.:x-3>0=>x>3
x-1>0=>x>1
2log₅(x-3)=log₅(x-1)<=>log₅(x-3)²-log₅(x-1)=0<=>log₅[(x-3)²/(x-1)]=0=>(x-3)²/(x-1)=5⁰<=>x²+9-6x=x-1<=>x²-7x+10=0
Δ=49-40=9
x₁=(7+3)/2=5
x₂=(7-3)/2=2 nu verifica C.E. deci nu convine
x=x₁=5=>S={5}