標題:
解指數定律方程一問
請計算以下題目,要有步驟 3^(x+1)+3^(x)-36=0 5^(x+1)-5x-100=0 9^(x)+9^(x-1)-10=0
最佳解答:
3^(x+1)+3^(x)-36=0 3(3^x) + 3^x = 36 (3+1)(3^x) = 36 3^x = 9 3^x = 3^2 x = 2 5^(x+1)-5x-100=0 5(5^x) - 5^x = 100 (5-1)(5^x) = 100 5^x = 25 5^x = 5^2 x = 2 9^(x)+9^(x-1)-10=0 9[9^(x-1)] + 9^(x-1) = 10 (9+1)(9^(x-1)) = 10 9^(x-1) = 1 9^(x-1) = 9^0 x-1 = 0 x = 1
其他解答:
3^(x+1)+3^(x)-36=0 3(3^x) + 3^x = 36 (3+1)(3^x) = 36 3^x = 9 3^x = 3^2 x = 2 5^(x+1)-5x-100=0 5(5^x) - 5^x = 100 (5-1)(5^x) = 100 5^x = 25 5^x = 5^2 x = 2 9^(x)+9^(x-1)-10=0 9[9^(x-1)] + 9^(x-1) = 10 (9+1)(9^(x-1)) = 10 9^(x-1) = 1 9^(x-1) = 9^0 x-1 = 0 x = 1
解指數定律方程一問
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發問:請計算以下題目,要有步驟 3^(x+1)+3^(x)-36=0 5^(x+1)-5x-100=0 9^(x)+9^(x-1)-10=0
最佳解答:
3^(x+1)+3^(x)-36=0 3(3^x) + 3^x = 36 (3+1)(3^x) = 36 3^x = 9 3^x = 3^2 x = 2 5^(x+1)-5x-100=0 5(5^x) - 5^x = 100 (5-1)(5^x) = 100 5^x = 25 5^x = 5^2 x = 2 9^(x)+9^(x-1)-10=0 9[9^(x-1)] + 9^(x-1) = 10 (9+1)(9^(x-1)) = 10 9^(x-1) = 1 9^(x-1) = 9^0 x-1 = 0 x = 1
其他解答:
3^(x+1)+3^(x)-36=0 3(3^x) + 3^x = 36 (3+1)(3^x) = 36 3^x = 9 3^x = 3^2 x = 2 5^(x+1)-5x-100=0 5(5^x) - 5^x = 100 (5-1)(5^x) = 100 5^x = 25 5^x = 5^2 x = 2 9^(x)+9^(x-1)-10=0 9[9^(x-1)] + 9^(x-1) = 10 (9+1)(9^(x-1)) = 10 9^(x-1) = 1 9^(x-1) = 9^0 x-1 = 0 x = 1
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