已知2^a×5^b=2^c×5^d=10,求证(a-1)(d-1)=(b-1)(c-1)(急~)

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已知2^a×5^b=2^c×5^d=10,求证(a-1)(d-1)=(b-1)(c-1)(急~)已知2^a×5^b=2^c×5^d=10,求证(a-1)(d-1)=(b-1)(c-1)(急~)已知2^

已知2^a×5^b=2^c×5^d=10,求证(a-1)(d-1)=(b-1)(c-1)(急~)
已知2^a×5^b=2^c×5^d=10,求证(a-1)(d-1)=(b-1)(c-1)(急~)

已知2^a×5^b=2^c×5^d=10,求证(a-1)(d-1)=(b-1)(c-1)(急~)
2^a×5^b=2^c×5^d=10
除以10
2^a×5^b/10=2^c×5^d/10=1
(2^a/2)×(5^b/5)=(2^c/2)×(5^d/5)=1
2^(a-1)×5^(b-1)=2^(c-1)×5^(d-1)=1
取对数
(a-1)lg2+(b-1)lg5=(c-1)lg2+(d-1)lg5=0
(a-1)lg2+(b-1)lg5=0
lg2/lg5=-(b-1)/(a-1)
=(c-1)lg2+(d-1)lg5=0
lg2/lg5=-(d-1)/(c-1)
所以-(b-1)/(a-1)=-(d-1)/(c-1)
(a-1)(d-1)=(b-1)(c-1)