Question

如果方程loga(x-5)/(x+5)=1+loga(x-3)有实根 求实数a的取值

Answer 1

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

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