Please solve e^(4x+1)=7?
回答 (8)
2009-12-20 10:55 am
✔ 最佳答案
e^(4x+1)=7lne^(4x+1)=ln(7)
4x+1=ln(7)
4x=ln(7)-1
....ln(7)-1
x=-----------
........4
x≈0.2364 answer//
2009-12-20 6:33 pm
4x + 1 = ln 7
4x = ln 7 - 1
x = ( ln 7 - 1 ) / 4
4x = ln 7 - 1
x = ( ln 7 - 1 ) / 4
2009-12-20 7:55 pm
To solve this you need to insert ln into both sides of equation. You will get 4x+1=ln7
Solve for x and you will get 0.2365
Solve for x and you will get 0.2365
2009-12-20 6:48 pm
x=0.2365
2009-12-20 6:45 pm
e^(4x + 1) = 7
4x + 1 = ln(7)
4x = ln(7) - 1
x = [ln(7) - 1]/4
4x + 1 = ln(7)
4x = ln(7) - 1
x = [ln(7) - 1]/4
2009-12-20 6:24 pm
e^(4x+1)=7
4x+1=ln7
4x+1=1.946
4x=1.946-1
4x=0.946
x=0.2365
4x+1=ln7
4x+1=1.946
4x=1.946-1
4x=0.946
x=0.2365
2009-12-20 6:17 pm
You should begin by taking the natural log of both sides.
ln(e^(4x+1) = ln(7))
4x+1 = 1.95
The rest is just simple algebra
4x = 0.95
x = .2375
ln(e^(4x+1) = ln(7))
4x+1 = 1.95
The rest is just simple algebra
4x = 0.95
x = .2375
2009-12-20 6:17 pm
e^(4x+1)=7
4x+1 = ln7
4x = ln7-1
x = (ln7-1)/4
x = (1.945-1)/4
x = 0.236
4x+1 = ln7
4x = ln7-1
x = (ln7-1)/4
x = (1.945-1)/4
x = 0.236
參考: ..
收錄日期: 2021-05-01 12:54:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091220021238AA1yb4y