更新时间2018-08-11 13:40:44
alog(b,c)-blog(a,c)
=algc/lgb-blgc/lga
=(algalgc)/(lgalgb)-(blgblgc)/(lgalgb)
=(algalgc-blgblgc)/(lgalgb)
=[(alga-blgb)lgc]/(lgalgb)①
∵a>b>1
lga>0,lgb>0②
∴alga-blgb>0③
∵0<c<1
∴lgc<0④
②③④代入①得
(正数*负数)/(正数*正数)
=负数/正数
=负数
<0
即:alog(b,c)-blog(a,c)<0
∴alog(b,c)<blog(a,c)
因为alog(b)c=log(b)c^a,blog(a)c=log(a)c^b
通过作图,知log(b)c^a<log(b)c^b,log(b)c^b<log(a)c^b
所以log(b)c^a<log(a)c^b
即alog(b)c<blog(a)c