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