已知abc 均为自然数 2的A次方 等于3的B次方 等于6的C次方求证ab-bc=ac

令2^A=3^B=6^C=D
则A=logD/log2;B=logD/log3;C=logD/log6
则ab-bc=（logD/log2）*（logD/log3）-（logD/log3）*（logD/log6）
=logD*logD（log6-log2）/(log2*log3*log6)
=logD*logD*(log(6/2))/(log2*log3*log6)
=logD*logD*log3/(log2*log3*log6)
=logD*logD/(log2*log6)
=(logD/log2)*(logD/log6)
=AC