log(2)3=a,则log(6)4=(

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log(2)3=a,则log(6)4=(
log(2)3=a,则log(6)4=(

log(2)3=a,则log(6)4=(
换底公式
a=lg3/lg2
lg3=alg2
所以原式= lg4/lg6
=2lg2/(lg2+lg3)
=2lg2/(lg2+alg2)
=2/(1+a)

由log(2)3=a得到log(3)2=1/a,log(6)4=log(2×3)2^2=2[log(2)2+log(3)2]=2(a+1/a)

log<2>3=a ===> lg3/lg2=a
===> lg3=a*lg2
log<6>4=lg(2^2)/lg(2*3)=(2lg2)/(lg2+lg3)
=(2lg2)/(lg2+a*lg2)
=(2lg2)/[(a+1)*lg2]
=2/(a+1)

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