
ㄱ. {f(5a)}2=(log55a)2=(log5a−1)2
{f(a5)}2=(log5a5)2=(1−log5a)2
∴ 참
ㄴ. f(a+1)−f(a)=log5(a+1)−log5a=log5(1+a1)
f(a+2)−f(a+1)
=log5(a+2)−log5(a+1)
=log5(1+a+11)
1+a1>1+a+11이므로
log5(1+a1)>log5(1+a+11) ∴ 참
ㄷ. f−1(x)=5x
f(a)<f(b)이면 a<b
a<b이면 f−1(a)<f−1(b) ∴ 참 따라서, ㄱ, ㄴ, ㄷ 모두 옳다.