先介绍换底公式：log(a)b=lgb/lga
证明：设log(a)b=t
则a^t=b,两边取以10为底的对数
lga^t=lgb
tlga=lgb
所以t=lgb/lga
所以log(a)b=lgb/lga
＝＝＝＝＝＝＝＝＝＝＝
log(8)9=a ==> lg9/lg8=a ==> 2lg3/(3lg2)=a ==> lg3=(3a/2)lg2
log(2)5=b ==> lg5/lg2=b ==> (lg10-lg2)/lg2=b ==> lg2=1/(b+1)
由此可得,lg3=3a/[2(b+1)]