e^2x-4e^x+3=0 help please correct me if I'm wrong?
2020-08-14 6:10 pm
"x - In(3)" is my answer and says that the answer should be in decimal form so it should be "1.09"? and says that round off the answer to 1 decimal place so it could be "1.1"?
回答 (3)
2020-08-14 6:23 pm
e²ˣ - 4eˣ + 3 = 0
(eˣ)² - 4eˣ + 3 = 0
(eˣ - 1)(eˣ - 3) = 0
eˣ = 1 or eˣ = 3
x = ln(1) or x = ln(3)
x = 0.0 or x = 1.1
x has two roots: 0.0 and 1.1
(eˣ)² - 4eˣ + 3 = 0
(eˣ - 1)(eˣ - 3) = 0
eˣ = 1 or eˣ = 3
x = ln(1) or x = ln(3)
x = 0.0 or x = 1.1
x has two roots: 0.0 and 1.1
2020-08-14 10:43 pm
e^(2x) - 4e^x + 3 = 0
Let u = e^x, so we have:
u^2 - 4u + 3 = 0
u^2 - u - 3u + 3 = 0
u(u - 1) - 3(u - 1) = 0
(u - 3)(u - 1) = 0
u = 3 or u = 1
e^x = 3 or e^x = 1
Exact answers (complex):
x = ln(3) + 2*i*pi*n, for any integer n, or
x = ln(1) + 2*i*pi*n = 2*i*pi*n, for any integer n
Approximate answers (complex):
x =~ 1.098612 + 6.283185in, for any integer n, or
x =~ 6.283185in, for any integer n
Exact answers (real):
x = ln(3) or x = 0
Approximate answers (real):
x =~ 1.098612 or x = 0
Let u = e^x, so we have:
u^2 - 4u + 3 = 0
u^2 - u - 3u + 3 = 0
u(u - 1) - 3(u - 1) = 0
(u - 3)(u - 1) = 0
u = 3 or u = 1
e^x = 3 or e^x = 1
Exact answers (complex):
x = ln(3) + 2*i*pi*n, for any integer n, or
x = ln(1) + 2*i*pi*n = 2*i*pi*n, for any integer n
Approximate answers (complex):
x =~ 1.098612 + 6.283185in, for any integer n, or
x =~ 6.283185in, for any integer n
Exact answers (real):
x = ln(3) or x = 0
Approximate answers (real):
x =~ 1.098612 or x = 0
2020-08-14 6:21 pm
e^(2x) - 4e^(x) + 3 = 0
(e^(x) - 1)(e^(x) - 3) = 0
e^(x) - 1 = 0, e^(x) - 3 = 0
e^(x) = 1, e^(x) = 3
x = ln(1), x = ln(3)
x = 0, x = 1.098 or 1.1
(e^(x) - 1)(e^(x) - 3) = 0
e^(x) - 1 = 0, e^(x) - 3 = 0
e^(x) = 1, e^(x) = 3
x = ln(1), x = ln(3)
x = 0, x = 1.098 or 1.1
收錄日期: 2021-04-24 08:03:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200814101052AA4MrNx