2^x - 2^(3-x) = 7
(find real roots)
Maths help
2010-09-27 1:33 am
回答 (2)
2010-09-27 1:40 am
✔ 最佳答案
2^x - 2^(3-x) = 72^x - (2^3)(2^-x) = 7(2^x) [ 2^x - (2^3)(2^-x) ] = 7 (2^x)(2^x)(2^x) - (2^3)(2^-x)(2^x) = 7 (2^x)(2^x)^2 - (2^3)(2^0) = 7(2^x)(2^x)^2 - 8 = 7(2^x)(2^x)^2 - 7(2^x) - 8 = 0(2^x - 8)(2^x + 1) = 02^x = 8 or 2^x = - 1(Rejected)2^x = 82^x = 2^3x = 3
2010-09-27 1:49 am
2^x - 2^(3-x) = 7
2^x - (2^3)(2^-x) = 7
(2^x)(2^x) - (2^3) = 7 (2^x)
(2^x)^2 - 7(2^x)-8=0
let 2^x =y
y^2-7y-8=0
(y-8)(y+1)=0
y=8 or -1
2^x = 8 or 2^x = - 1(rej)
2^x = 8
x = 3
2^x - (2^3)(2^-x) = 7
(2^x)(2^x) - (2^3) = 7 (2^x)
(2^x)^2 - 7(2^x)-8=0
let 2^x =y
y^2-7y-8=0
(y-8)(y+1)=0
y=8 or -1
2^x = 8 or 2^x = - 1(rej)
2^x = 8
x = 3
收錄日期: 2021-04-21 22:17:28
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