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已知2^a乘5^b=2^c乘5^d=10,求证:(a-1)(d-1)=(b-1)(c-1)

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已知2^a乘5^b=2^c乘5^d=10,求证:(a-1)(d-1)=(b-1)(c-1)
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已知2^a乘5^b=2^c乘5^d=10,求证:(a-1)(d-1)=(b-1)(c-1)
2^a×5^b=2^c×5^d=10
两边除以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
2^(a-1)×5^(b-1)=1
[2^(a-1)×5^(b-1)]^(d-1)=1^(d-1)
2^(a-1)(d-1)×5^(b-1)(d-1)=1
2^(c-1)×5^(d-1)=1
[2^(c-1)×5^(d-1)]^(a-1)=1^(a-1)
2^(a-1)(c-1) ×5^(d-1)(a-1)=1
相乘
2^(a-1)(d-1)×5^(b-1)(d-1)×2^(a-1)(c-1) ×5^(d-1)(a-1)=1
10^(a-1)(d-1)×5^(b-1)(d-1)×2^(a-1)(c-1)=1
2^(a-1)×5^(b-1)=2^(c-1)×5^(d-1)=1
2^(a-1)×5^(b-1)=1
[2^(a-1)×5^(b-1)]^(c-1)=1^(c-1)
2^(a-1)(c-1)×5^(c-1)(d-1)=1
2^(c-1)×5^(d-1)=1
[2^(c-1)×5^(d-1)]^(b-1)=1^(b-1)
2^(c-1)(b-1) ×5^(d-1)(b-1)=1
相乘
2^(a-1)(c-1)×5^(c-1)(d-1)×2^(c-1)(b-1) ×5^(d-1)(b-1)=1
10^(c-1)(b-1)×5^(b-1)(d-1)×2^(a-1)(c-1)=1
所以10^(a-1)(d-1)×5^(b-1)(d-1)×2^(a-1)(c-1)=10^(c-1)(b-1)×5^(b-1)(d-1)×2^(a-1)(c-1)
所以10^(a-1)(d-1)=10^(c-1)(b-1)
(a-1)(d-1)=(c-1)(b-1)