An Interesting Exponential Equation

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  • Опубліковано 13 січ 2025

КОМЕНТАРІ • 4

  • @spelunkerd
    @spelunkerd День тому

    Even when I get the answer, I often learn a little more from your discussion. In this case it was the short hand method of long division, and the conjugate complex answer.

  • @Don-Ensley
    @Don-Ensley День тому

    problem
    3ˣ^² • 9ˣ = 27 ᵛˣ̅
    Put In terms of powers of 3.
    3ˣ^² • 3² ˣ = 3 ³ ᵛˣ̅
    3 ⁽ ˣ^² ⁺ ²ˣ ⁾ = 3 ³ ᵛˣ̅
    Each side has the same base of 3. Set the exponents equal.
    x² + 2 x = 3 √x
    Let
    y = √x
    , the positive valued root.
    x = y²
    y⁴ + 2 y²- 3 y = 0
    y ( y³ + 2y -3) = 0
    Use the 0 product property.
    The first term gives us
    y = 0
    , for which x = y² = 0.
    The second term says
    y³ + 2y -3 = 0
    , which has a coefficient sum of 0.
    y = 1 is a root.
    , for which x = y² = 1.
    Factor out y-1.
    (y-1) y² +(y-1) y + 3(y-1) = 0
    (y-1)(y² + y + 3) = 0
    Apply the 0 product property.
    y² + y + 3 = 0
    y = ( -1 ± i √11 ) / 2
    , for which
    x = y² = (-5 ± i √11 ) / 2
    answer
    x ∈ { 0, 1,
    (-5 - i √11 ) / 2,
    (-5 + i √11 ) / 2 }

  • @rob876
    @rob876 День тому +1

    x^2 + 2x = 3√x
    x = 0
    or
    let u = √x
    u^3 + 2u - 3 = 0
    u^2(u - 1) + u(u - 1) + 3(u - 1) = 0
    (u - 1)(u^2 + u + 3) = 0
    u = 1
    => x = 1
    or
    u = (-1 ± i√11)/2
    => x = ((-1 ± i√11)/2)^2
    => x = (-5 ± i√11)/2

  • @prollysine
    @prollysine День тому

    let u=Vx , x=u^2 , u^4+2u^2=3u , /: u , u^3+/-u^2+2u-3=0 , (u-1)(u^2+u+3)=0 , u=1 . x=u^2 , x=1^2 , x=1 ,
    1 -1 / u=(-1+/-i*V11)/2 , / ,
    1 -1 test , 3^(1^2)*9^1=27 , 27^(V1)=27 , same , OK ,
    3 -3