e^2x − 9^ex + 8 = 0
Been working this problem for a while now and can't seem to figure it out.
更新1:
Yeah my mistake the equation is actually e^2x − 9e^x + 8 = 0
Natural Logs Question?
2015-09-03 11:32 pm
Solve the equation. Give exact answers. Do not use your calculator.
e^2x − 9^ex + 8 = 0
Been working this problem for a while now and can't seem to figure it out.
e^2x − 9^ex + 8 = 0
Been working this problem for a while now and can't seem to figure it out.
回答 (3)
2015-09-03 11:40 pm
The equation should be e^2x − 9e^x + 8 = 0 instead
Put u = e^x
e^2x − 9e^x + 8 = 0
(e^x)^2 − 9e^x + 8 = 0
u^2 - 9u + 8 = 0
(u - 1)(u - 8) = 0
u = 1 or u = 8
e^x = 1 or e^x = 8
log(e^x) = log(1) or log(e^x) = log(8)
x log(e) = 0 or x ln(e) = ln(8)
x = 0 or x = ln(8)
Put u = e^x
e^2x − 9e^x + 8 = 0
(e^x)^2 − 9e^x + 8 = 0
u^2 - 9u + 8 = 0
(u - 1)(u - 8) = 0
u = 1 or u = 8
e^x = 1 or e^x = 8
log(e^x) = log(1) or log(e^x) = log(8)
x log(e) = 0 or x ln(e) = ln(8)
x = 0 or x = ln(8)
2015-09-03 11:37 pm
Assuming it is
e^(2x)-9e^x+8 = 0
(e^x)^2-9e^x+8 = 0
(e^x-8)(e^x-1) = 0
e^x = 1 or 8
x = ln 1 or ln 8
e^(2x)-9e^x+8 = 0
(e^x)^2-9e^x+8 = 0
(e^x-8)(e^x-1) = 0
e^x = 1 or 8
x = ln 1 or ln 8
2015-09-03 11:39 pm
u = e^x
u^2 - 9u + 8 = 0
(u-8)(u-1)
u = {8,1}
e^x = 8, x = ln 8
e^x = 1, x = 0
u^2 - 9u + 8 = 0
(u-8)(u-1)
u = {8,1}
e^x = 8, x = ln 8
e^x = 1, x = 0
收錄日期: 2021-04-18 00:10:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150903153225AAtctzv