Natural Logs Question?

👨🏻‍🏫 學術台數學

[匿名]

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.

更新1:

Yeah my mistake the equation is actually e^2x − 9e^x + 8 = 0

回答 (3)

不用客氣

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)

👍 1 👎 0

用戶57799

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

👍 1 👎 0

[匿名]

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

👍 0 👎 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

檢視 Wayback Machine 備份