若a>0,a≠1;b>0,b≠1;c>0,c≠1有 a~(log_cb)=b~(log_ca) 证 log_cb·log_ca=log_cb·log_calog_ca~(log_cb)=log_cb~(log_ca)a~(log_cb)=b~(log_ca)利用这个等式可证明下面这道题设 log_(2a)a=x,log_(3a)2a=y 求证 2~(1-xy=3~(y-xy) 证明左=2~(1-log_(2a)alog_(3a)n2a) =2~(log_(3a)3a-log_(3a)a=2~(log_(3a)3 右:3~(log_(3a)2a-log_(2a)alog_(3a)2a =3~(log_(3a)2a-log_(3a)a=3~(log_(3a)2 ∵2~(log_(3a)~2-3~(log_(3a)2) (据a~(log_cb=b~(log_ca) ∴2~(1-xy)=3~(y-xy) 成立这个等式时我们解有关数学题会大有帮助。为了使有限的版面刊登更多的作品,使更多的作者能将自已在教学中的 If a>0,a≠1;b>0,b≠1;c>0,c≠1 has a~(log_cb)=b~(log_ca) certificate log_cb·log_ca=log_cb·log_calog_ca~(log_cb) =log_cb~(log_ca)a~(log_cb)=b~(log_ca) Using this equation, we can prove that the following problem is set to log_(2a)a=x, log_(3a)2a=y. Proof 2~(1) Xy=3~(y-xy) proves that left=2~(1-log_(2a)alog_(3a)n2a)=2~(log_(3a)3a-log_(3a)a=2~(log_(3a) 3 Right: 3~(log_(3a)2a-log_(2a)alog_(3a)2a=3~(log_(3a)2a-log_(3a)a=3~(log_(3a)2~2~(log_ (3a)~2-3~(log_(3a)2) (according to a~(log_cb=b~(log_ca)~2~(1-xy)=3~(y-xy)) Questions about mathematics can be of great help in order to allow more pages to be published, so that more authors will be able to