就当真数是(x-5)/(x+5)
loga(x-5)/(x+5)=1+loga(x-3)
loga(x-5)/(x+5)=loga[a(x-3)]
(x-5)/(x+5)=a(x-3)
x-5=a(x-3)(x+5)
ax^2+(2a-1)x+(5-15a)=0
要使方程有实根,则Δ≥0
(2a-1)^2-4a(5-15a)≥0
64a^2-24a+1≥0
0