solve for the equation 10+5e^2x=17?
回答 (4)
2009-04-29 8:15 am
✔ 最佳答案
10+5e^2x=175e^2x=7
e^2x=7/5
ln(7/5)=2x
x=1/2.ln(7/5)
thats about 0.168236118 if you had to round it
2009-04-29 9:57 am
10 + 5e^(2x) = 17
5e^(2x) = 17 - 10
e^(2x) = 7/5
ln[e^(2x)] = ln(7/5)
(2x)ln(e) = ln(7/5)
(2x)(1) = ln(7/5)
x = [ln(7/5)]/2
5e^(2x) = 17 - 10
e^(2x) = 7/5
ln[e^(2x)] = ln(7/5)
(2x)ln(e) = ln(7/5)
(2x)(1) = ln(7/5)
x = [ln(7/5)]/2
2009-04-29 8:24 am
0.168236118310606465
2009-04-29 8:15 am
e^2x=7/5
2xlne=ln7/5
x=0.18236118
2xlne=ln7/5
x=0.18236118
收錄日期: 2021-05-01 12:24:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090429000611AA7Y0RP