已知log14底7=a ,log14底5=b ,则log35底28=

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已知log14底7=a ,log14底5=b ,则log35底28=
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已知log14底7=a ,log14底5=b ,则log35底28=
已知log14底7=a ,log14底5=b ,则log35底28=

已知log14底7=a ,log14底5=b ,则log35底28=
log3528=log1428/log1435
=(log1414+log142)/(log147+log145)
=(1+log14 (14/7))/(a+b)
=(1+log1414-log147)/(a+b)
=(1+1-a)/(a+b)
=(2-a)/(a+b)
PS:怕你不明白,下面有点解释(以14为底)
因为Log147+log142=1,所以log142=1-a
因为log147+log145=log1435,所以log1435=a+b
Log3528=log1428/log1435=(1+log142)/(a+b)=(2-a)/(a+b)