Solve for x,2e^2x-8e^x+6=0?

8 Answers

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  • 4 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) }

  • Philip
    Lv 6
    1 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).

  • 1 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.

  • 1 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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  • 1 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....

  • 1 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)

  • Anonymous
    1 month ago

    Here you will understand better.

    Attachment image
  • 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)

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