1. If 3x^-1/2 = 4 , where x > 0 , then x =
2. If 7^x + 7^x+1 = 392 , then x =
please give a detail explanation!! Thank you
about some math questions!!!
2011-11-25 4:53 am
回答 (3)
2011-11-25 5:11 am
✔ 最佳答案
1) 3x^-1/2 = 41/(root x) =4/3
root x =3/4
x=(3/4)^2 or x=-(3/4)^2
x=9/16 or x=-9/16(rejected)
Therefore, x=9/16
2) 7^x + 7^x+1 = 392
7^x(1+7) = 392
7^x = 49
7^x = 7^2
x = 2
參考: me
2011-11-25 5:14 am
1.
3x^(-1/2)=4
3/x^(1/2)=4
3/4=x^(1/2)
x=9/16
2.
7^x+7^(x+1)=392
7^x+7^x*7=392
8*7^x=392
7^x=49
x=2
3x^(-1/2)=4
3/x^(1/2)=4
3/4=x^(1/2)
x=9/16
2.
7^x+7^(x+1)=392
7^x+7^x*7=392
8*7^x=392
7^x=49
x=2
2011-11-25 5:14 am
您好，我是 lop，高興能解答您的問題。
Maybe you have learned log. , but I still try to not use log. .
(1)
3x^(-1/2) = 4
3/√x = 4
3 = 4√x
3/4 = √x
(√x)^2 = (3/4)^2
x = 9/16
(2)
7^x + 7^(x+1) = 392
7^x + 7^x × 7^1 = 392
7^x + 7(7^x) = 392
8(7^x) = 392
7^x = 392 ÷ 8
7^x = 49
Taking log. to the base 7 of the both sides or
7^x = 49
7^x = 7^2
x = 2
Maybe you have learned log. , but I still try to not use log. .
(1)
3x^(-1/2) = 4
3/√x = 4
3 = 4√x
3/4 = √x
(√x)^2 = (3/4)^2
x = 9/16
(2)
7^x + 7^(x+1) = 392
7^x + 7^x × 7^1 = 392
7^x + 7(7^x) = 392
8(7^x) = 392
7^x = 392 ÷ 8
7^x = 49
Taking log. to the base 7 of the both sides or
7^x = 49
7^x = 7^2
x = 2
參考: Hope I Can Help You ^_^ ( From me )
收錄日期: 2021-04-16 13:45:20
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