Question

设f（x）=(log2^2 x )+(5log2 x)+1 若f（a）=f（b）=0 且a≠b 求 ab

Answer 1

f（a）=(log2^2 a )+(5log2 a)+1
f（b）=(log2^2 b )+(5log2 b)+1
f(a)=f(b)
所以 (log2^2 a )+(5log2 a)+1 = (log2^2 b )+(5log2 b)+1
所以（因式分解）（log2 a + log2 b + 5)(log2 a - log2 b)=0
a≠b ,所以 log2 a ≠ log2 b
所以 log2 a + log2 b + 5 =0
log2 ab = -5
ab = 2^(-5) = 1/32
其实 f（a）=0，f（b）= 0的条件有点多余。