已知log以2为底3的对数=a,log以3为底7的对数=b,求log以14为底56的对数

问题描述:

已知log以2为底3的对数=a,log以3为底7的对数=b,求log以14为底56的对数

log2(3)=a
log3(7)=b
log4(56)=?

log2(7)=log2(3)×log3(7)=ab
log4(56)
=log2²(7×2³)
=(1/2)[log2(7)+log2(2³)]
=(1/2)[ab+3]
=ab/3+3/2

哪里有问题请指出,没问题请采纳~~~~~~~~~~~~~~

log2 3=a
log3 2=1/a
log2 7=log3 7/log3 2=ab
log14 56
=1+log14 4
=1+2log14 2
=1+(2/log2 14)
=1+[2/(1+log2 7)]
=1+[2/(1+ab)]

答:
a=log2(3),b=log3(7)=log2(7)/log2(3)
所以:ab=log2(7)
log14(56)
=log2(7*8)/log2(14)
=[log2(7)+3]/[log2(7)+1]
=(ab+3)/(ab+1)
本题考查的就是对数函数的换底公式