2^(a-1)*5^(b-1)=1
=>(a-1)lg2+(b-1)lg5=0
同理(c-1)lg2+(d-1)lg5=0
=>(a-1)(d-1)=(b-1)(c-1)

2^(a-1)*5^(b-1)=1
2^(c-1)*5^[(b-1)(c-1)/(a-1)]=1^[(c-1)/(a-1)]=1=2^(c-1)*5^(d-1)
比较得(b-1)(c-1)/(a-1)=(d-1) 懂了吗