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### 8 Answers

- la consoleLv 74 weeks ago
2e^(2x) - 8e^(x) + 6 = 0 → you can simplify by 2

e^(2x) - 4e^(x) + 3 = 0

e^(x + x) - 4e^(x) + 3 = 0 → you know that: x^(a + b) = (x^a) * (x^b)

[e^(x) * e^(x)] - 4e^(x) + 3 = 0 → let: X = e^(x) where: X > 0

[X * X] - 4X + 3 = 0

X² - 4X + 3 = 0

X² - 4X + 4 - 1 = 0

X² - 4X + 4 = 1

(X - 2)² = 1

X - 2 = ± 1

X = 2 ± 1

First case: X = 3

e^(x) = 3

x = Ln(3)

Second case: X = 1

e^(x) = 1

x = Ln(1)

x = 0

→ Solution = { 0 ; Ln(3) }

- PhilipLv 61 month ago
2e^2x -8e^x +6 = 0;

Put u = e^x. Then 2u^2-8u+6 = 0, ie., u^2-4u+3 = 0, ie., (u-1)(u-3) = 0, ie.,;

e^x = 1...(i) or e^x = 3...(ii).;

For (i) holding, x = ln(1) = 0.;

For (ii) holding, x = ln(3).

- PinkgreenLv 71 month ago
2e^(2x)-8e^x+6=0

=>

e^(2x)-4e^x+3=0

=>

(e^x-3)(e^x-1)=0

e^x=3=>x=ln(3).

e^x=1=>x=ln(1)=0.

Either x=ln(3) or x=0 is the solutions.

- Iggy RockoLv 71 month ago
Assuming 2e^2x should be 2e^(2x),

2e^(2x) - 8e^x + 6 = 0

e^(2x) - 4e^x + 3 = 0

(e^x - 3)(e^x - 1) = 0

e^x = 3 or e^x = 1

x = ln3 or x = 0

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- Engr. RonaldLv 71 month ago
2e^2x-8e^x+6=0

[2e^(x) - 6][e^(x) - 1] = 0

2e^(x) - 6 = 0, e^(x) - 1 = 0

e^(x) = 3 , e^(x) = 1

x = ln(3), x = ln(1)

x = ln(3) or 0....

- micatkieLv 61 month ago
2e²ˣ - 8eˣ + 6 = 0

e²ˣ - 4eˣ + 3 = 0

(eˣ)² - 4(eˣ) + 3 = 0

(eˣ - 1)(eˣ - 3) = 0

eˣ = 1 or eˣ = 3

x = ln(1) or x = ln(3)

x = 0 or x = ln(3)

- Anonymous1 month ago
Here you will understand better.

- 1 month ago
e^x = k

2k^2 - 8k + 6 = 0

k^2 - 4k + 3 = 0

(k - 3) * (k - 1) = 0

k = 1 , 3

e^x = 1 , 3

x = ln(1) , ln(3)

x = 0 , ln(3)