Question

已知2^a3^b=2^c3^d,求证,（a-1)(d-1)=(b-1)(c-1)

Answer 1

令2'a3'b =2'c3'd = T两边同时去以10为底的对数 lg(2^a*3^b)=T——>alg2+blg3=T——>b=(T-alg2)/lg3 lg(2^c*3^d)=T——>clg2+dlg3=T——>d=(T-clg2)/lg3 则 (a-1)(d-1)=(a-1)((1-clg2-lg3)/lg3) =(a-1)(lg2-clg2)/lg3=(a-1)(1-c)(1-lg2)/lg3 =(a-1)(1-c) (b-1)(c-1)=(1-alg2-lg3)(c-1)/lg3 =(lg2-alg2)(c-1)/lg3=(1-a)(c-1)lg3/lg3 =(1-a)(c-1) 显然(a-1)(d-1)=(b-1)(c-1) 等式成立.