An Interesting Exponential Equation

[@spelunkerd]

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]

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]

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]

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