1. 9(3^x+1)=27^2x
2. 2^(x+2)+2^x=10
指數方程問題
2007-04-17 1:37 am
回答 (2)
2007-04-17 1:40 am
✔ 最佳答案
1.9(3^x+1)=27^2x
3^2 (3^x + 1) = (3 ^ 3) ^ 2x
3^2x + 9 = 3 ^ 6x
3 ^ 4x = 9
3^ 4x = 3 ^ 2
4x = 2
x = 1 / 2
2.
2^(x+2)+2^x=10
2 ^x (4 + 1) = 10
2^ x (5) = 10
2^x = 2
x = 1
2007-04-17 2:21 am
1.
9(3^x+1)=27^2x
3^2 (3^x + 1) = (3 ^ 3) ^ 2x
3^2x + 9 = 3 ^ 6x
3 ^ 4x = 9
3^ 4x = 3 ^ 2
4x = 2
x = 1 / 2
2.
2^(x+2)+2^x=10
2 ^x (4 + 1) = 10
2^ x (5) = 10
2^x = 2
x = 1
9(3^x+1)=27^2x
3^2 (3^x + 1) = (3 ^ 3) ^ 2x
3^2x + 9 = 3 ^ 6x
3 ^ 4x = 9
3^ 4x = 3 ^ 2
4x = 2
x = 1 / 2
2.
2^(x+2)+2^x=10
2 ^x (4 + 1) = 10
2^ x (5) = 10
2^x = 2
x = 1
參考: me
收錄日期: 2021-04-14 18:52:05
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