2^a×3^b=2^c×3^d=6,求证:(a-1)×(d-1)=(c-1)×(b-1)

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2^a×3^b=2^c×3^d=6,求证:(a-1)×(d-1)=(c-1)×(b-1)

2^a×3^b=2^c×3^d=6,求证:(a-1)×(d-1)=(c-1)×(b-1)
2^a×3^b=2^c×3^d=6,求证:(a-1)×(d-1)=(c-1)×(b-1)

2^a×3^b=2^c×3^d=6,求证:(a-1)×(d-1)=(c-1)×(b-1)
证法1:
因为 2^a*3^b=2^c*3^d=6,
所以 2^(a-1)=6/(2*3^b)=3^(1-b),
2^(c-1)=6/(2*3^d)=3^(1-d).
所以 2^[(a-1)(c-1)]=3^[(1-b)(c-1)],
2^[(c-1)(a-1)]=3^[(1-d)(a-1)].
所以 3^[(1-b)(c-1)]=3^[(1-d)(a-1)].
所以 (a-1)*(d-1)=(c-1)*(b-1).
证法2:
因为 2^a*3^b=6,
取对数得
alg2+blg3=lg2+lg3.
即 (a-1)lg2+(b-1)lg3=0,(1)
同理 (c-1)lg2+(d-1)lg3=0.(2)
(1)*(c-1)-(2)*(a-1)得
(b-1)(c-1)lg3-(d-1)(a-1)lg3=0.
所以 (a-1)*(d-1)=(c-1)*(b-1).
= = = = = = =
注意:a,b,c,d可取1.所以(1)不能化为
lg2/lg3=(1-b)/(a-1).

证明:等式全部除以2*3得:
2^(a-1)*3^(b-1)=2^(c-1)*3^(d-1)=1
再取log得:
(a-1)log2+(b-1)log3=(c-1)log2+(d-1)log3=0
移项得:loga(a-1)/(c-1)-log3(b-1)/(d-1)=1-1=0
∴(a-1)/(c-1)=(b-1)/(d-1)=1
得证