Exponential Problem Trick that's Revolutionalizing Harvard University Entrance Exams

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Pass the Harvard University Entrance Exam with THIS Diophantine Problem Trick
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[@souzasilva5471]

Seja 3^x = a.
(a^-3^1/2)(a^2+a+4)=0, a = 3^1/2, logo x = 1/2 (Let 3^x = a.
(a^-3^1/2)(a^2+a+4)=0, a = 3^1/2, therefore x = 1/2)

[@shashankkatiha9439]

simply put sqrt(12) as 2sqrt(3). y^3-y=2sqrt(3). or y^3-y-2sqrt(3)=0. y=sqrt(3) is satisfying the equation here. as (sqrt(3))^3-sqrt(3)=3sqrt(3)-sqrt(3)=2sqrt(3). hence if y=sqrt(3)=3^1/2 then 3^x=3^1/2. which implies x=1/2. no need to go this long.

[@SGuerra]

É muito interessante que não haja a pesquisa de raízes racionais nas soluções de muitas questões fora do Brasil. Normalmente elas dão certo e facilitam muito a diminuição do grau do polinômio em questão. Fazer apenas fatoração pode virar uma cilada longa, muito longa. Brasil Outubro de 2024. It is very interesting that there is no research into rational roots in the solutions to many issues outside of Brazil. They usually work and make it much easier to reduce the degree of the polynomial in question. Just doing factoring can turn into a long, very long trap. Brazil October 2024.

[@kyintegralson9656]

Wasn't specified at the start that x is real. So case 2 can't be rejected, since it has complex x as solution. Unfortunately, this kind of inconsistency keeps getting repeated in this channel, no matter how many times it's been pointed out.
For case 2, y is given by
y=(-√3±i√5)/2=√2·e^(±iθ) where cosθ=-√(⅜) & sinθ=√(⅝) ⇒ θ≅.71π rad≅127.8°
x=log₃y=(½)log₃2±iθ/ln3.
Wasn't that bad, was it?

[@walterwen2975]

Harvard University Entrance Exam: 27ˣ - 3ˣ = √12; x =?
3ˣ > 0, No complex or imaginary value root for x
27ˣ - 3ˣ - √12 = 0, [(3ˣ)³ - 3√3)] - 3ˣ + √3 = 0, [(3ˣ)³ - (√3)³] - (3ˣ - √3) = 0
(3ˣ - √3)[(3ˣ)² + (√3)(3ˣ) + 3 - 1] = (3ˣ - √3)[(3ˣ)² + (√3)(3ˣ) + 2] = 0
(3ˣ)² + (√3)(3ˣ) + 2 > 0; 3ˣ - √3 = 0, 3ˣ = √3 = 3¹⸍², x = 1/2
Answer check:
x = 1/2, 3ˣ = √3: 27ˣ - 3ˣ = (3ˣ)[(3ˣ)² - 1] = (√3)[(√3)² - 1] = 2√3 = √12; Confirmed
Final answer:
x = 1/2

[@prollysine]

let u=3^x , then let k= u^2 , k^3-2k^2+k-12=0 , (k-3)(k^2+k+4)=0 , k=3 , k=u^2 , u=+/-V3 , / -V3 < 0 false / 3^x=V3 ,
x=logV3/log3 --> , x=(1/2 * log3)/log3 , x=1/2 , test , 27^(1/2) - 3^(1/2) = 2*V3 , 2*V3=V12 , OK ,