设logaC,logbC是方程x^2-3x+1=0的两根,求log(a/b) C的值
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发布时间：2024-10-20 13:44
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热心网友
时间：2024-11-14 16:09
解：由韦达定理，
log a (C)+log b (C)=3, (1)
log a (C)*log b (C)=1. (2)
令 log C (a)=u,log C (b)=v,则
log a (C)=1/u,
log b (C)=1/v.
分别代入(1),(2)得
1/u+1/v=3,
(1/u)*(1/v)=1.
即
u+v=3uv, (3)
uv=1. (4)
将(4)代入(1)得,
u+v=3. (5)
由(4),(5)得
(u-v)^2=(u+v)^2-4uv
=3^2-4*1
=5.
因此 u-v=正负根号（5).
即 log C (a/b)=正负根号(5).
故 log (a/b) C=1/[log C (a/b)]
=正负根号(5)/5.
解法二：
由韦达定理，
log a (C)+log b (C)=3, (1)
log a (C)*log b (C)=1. (2)
由(2)得，
log a (C)=1/log b (C)=log C (b),
log b (C)=1/log a (C)=log C (a).
代入(1),(2)得
log C (a)+log C (b)=3, (3)
log C (a)*lob C (b)=1. (4)
由(3),(4)得
(log C (a)-log C (b))^2
=(log C (a)+log C (b))^2-4*log C (a)*lob C (b)
=3^2-4*1
=5.
因此，log C (a)-log C (b)=正负根号（5).
即 log C (a/b)=正负根号（5).
故 log (a/b) C=1/[log C (a/b)]
=正负根号(5)/5.
热心网友
时间：2024-11-14 16:10
logaclogbc=1,logbc-logac=±√(logbc-logac)^2=±√(logbc+logac)^2-4logaclogbc=±√[(3/2)^2-4]=±5/2.
log(a/b) C=1/logc(a/b)=1/(logca-logcb)=1/(1/logac-1/logbc)=logaclogbc/(logbc-logac)=±5/2.
希望对你有帮助.
