Solve for x where loge(4+3x) = 2?
answer is 1/3 (e^2 - 4)
i wonder how to do it
Solve for x where loge(4+3x) = 2?
2011-06-19 11:03 am
回答 (3)
2011-06-19 11:08 am
✔ 最佳答案
ln (4+3x) = 2e^2 = (4+3x)
3x = e^2 - 4
x = (e^2 -4) /3
answer
參考: my brain (Prof TBT)
2011-06-19 6:09 pm
ln(4+3x) = 2
e^2=4+3x
e^2-4=3x
x=(e^2-4)/3
x=(1/3)(e^2-4)
e^2=4+3x
e^2-4=3x
x=(e^2-4)/3
x=(1/3)(e^2-4)
2011-06-19 6:08 pm
raise both sides to a power of "e":
4+3x = e^2 (e^loge cancels)
3x = e^2 - 4 (subtract 4 from both sides)
x = 1/3 (e^2 - 4) (divide both sides by 3)
4+3x = e^2 (e^loge cancels)
3x = e^2 - 4 (subtract 4 from both sides)
x = 1/3 (e^2 - 4) (divide both sides by 3)
收錄日期: 2021-05-01 00:52:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110619030335AAJzh0q