已知２＾ａ＊３＾ｂ＝２＾ｃ＊３＾ｄ＝６求證（ａ－１）（ｄ－１）＝（ｂ－１）（ｃ－１）

熱心網友

(a-1)/(c-1)=(a-c+c-1)/(c-1) =(a-c)/(c-1)+1 =(d-b)lg3/((c-1)lg2)+1 (2^a*3^b=2^c*3^d---(a-c)lg2=(d-b)lg3) =(d-b)/(1-d)+1 (2^c*3^d=6--c-1=(1-d)lg3/lg2) =(b-1)/(d-1)--(a-1)(d-1)=(b-1)(c-1)

熱心網友

已知：2^a*3^b＝2^c*3^d＝６,求證：(a-1)(d-1)=(b-1)(c-1)證明：2^a*3^b=6---alg2+blg3=lg6....(1)2^c*3^d=6---clg2+dlg3=lg6....(2)(1)*d-(2)*b:lg2=(d-b)lg6/(ad-bc)....(3)(2)*a-(1)*c:lg3=(a-c)lg6/(ad-bc)....(4)(3)+(4):(d-b)+(a-c)=(ad-bc)---ad-a-d=bc-b-c---(a-1)(d-1)=(b-1)(c-1)