设log27 12=a,则log6 16=

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设log27 12=a,则log6 16=

设log27 12=a,则log6 16=
设log27 12=a,则log6 16=

设log27 12=a,则log6 16=
log(12)27
把对数化成以2为底的对数,并且省略底数2
=log27/log12=log(3^3)/log(3*2^2)=3log3/(2+log3)
log(12)27=a--->3log3/(2+log3)=a--->log3=2a/(3-a)
log(6)16
=log(2^4)/log(2*3)
=4/(1+log3)
=4/[1+2a/(3-a)]
=4(3-a)/(3+a)

抱歉,我上初三。。

log12^27=3lg3/(2lg2+lg3)=a
lg2=(3lg3-a.lg3)/2a
log6^16=lg16/lg6=4lg2/(lg2+lg3)
=(12-4a)/(3+a)

log12^27=a,求log6^16.
a=lg27/lg12
=3lg3/(lg3+2lg2);
lg2/lg3
=(3-a)/ 2a,
lg3/lg2=2a/(3-a);
lg6/lg16
=(lg3+lg2)/4lg2
=1/4 (lg3/lg2+1)
=(3+a)/4(3-a)
lg16/lg6
=(12-4a)/(3+a).