（1）以知log√3^2=(1-
1)log(√3)2=(1-a)/a
--->log(3)2/(1/2)=2log(3)2=(1-a)/a
--->log(3)2=(1-a)/(2a)
log(12)3=log(3)3/log(3)(3*2^2)
=1/[1+2log(3)2]
=1/[1+2*(1-a)/(2a)]
=a/[a+(1-a)]
=a。
2)log(3)4=b--->2log(3)2=b--->log(3)2=b/2
log(6)7=a--->log(3)7/log(3)(2*3)=a
--->log(3)7/([+log(3)2]=a
--->log(3)7=a[1+log(3)2]
log(3)2=...全部
1)log(√3)2=(1-a)/a
--->log(3)2/(1/2)=2log(3)2=(1-a)/a
--->log(3)2=(1-a)/(2a)
log(12)3=log(3)3/log(3)(3*2^2)
=1/[1+2log(3)2]
=1/[1+2*(1-a)/(2a)]
=a/[a+(1-a)]
=a。
2)log(3)4=b--->2log(3)2=b--->log(3)2=b/2
log(6)7=a--->log(3)7/log(3)(2*3)=a
--->log(3)7/([+log(3)2]=a
--->log(3)7=a[1+log(3)2]
log(3)2=b/2--->log(3)7=a(1+b/2)
所以log(14)21=log(3)21/log(3)14
=log(3)(3*7)/log(3)(2*7)
=[1+log(3)7]/[log(3)2+log(3)7)]
=[1+a(1+b/2)]/[[b/2+a(1+b/2)]
=(ab+2a+2)/(ab+2a+b)。
收起
