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2005-12-14 10:58:19

• (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)

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  2005-12-14 10:58:19

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2005-12-13 17:11:04

• 已知：2^a*3^b＝2^c*3^d＝６,
  求证：(a-1)(d-1)=(b-1)(c-1)
  证明：
  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)
  

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  2005-12-13 17:11:04

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