A Nice Exponential Equation | Thailand Junior Math Olympiad

[@kassuskassus6263]

Let a=(2^2x+3)/4 and solve for a 128a^3-16a^2+8a-1=0. We'll get a=1/8. Thus, 2^2x+3=1/2, or 2^2x+4=1, so 2x+4=0, or x=-2.

[@user-kp2rd5qv8g]

Let t = 2^(2x+4)= 4^(x+2). Then, the given equation is 1/2 t^(1/2) = 1/2(2-t)^(1/3) > t^3 = (t-2)^2. t=1 is a solution and the only real solution.> 4^(x+2) = 1 > x=-2.

[@RyanLewis-Johnson-wq6xs]

Input
sqrt(1/8×2^(2×(-2) + 5)) = (1/4 (1 - 2^(2×(-2) + 3)))^(1/3)
Result
True
Left hand side
sqrt(1/8 2^(2 (-2) + 5)) = 1/2
Right hand side
(1/4 (1 - 2^(2 (-2) + 3)))^(1/3) = 1/2
Logarithmic form
1/2 log(4, 1/8 2^(2 (-2) + 5)) = 1/3 log(4, 1/4 (1 - 2^(2 (-2) + 3)))
X=-2

[@RyanLewis-Johnson-wq6xs]

I did it in my head.

[@user-ny6jf9is3t]

Αν y=2^χ>0 ,32y^3+8y^2-1=0 ,4(y-1/4)(8y^2+4y+1)=0,y=1/4=2^(-2), χ=-2 Δ

[@davidshen5916]

Y=2^（2X+3）， （Y/2）^（1/2）=（（1-Y）/4））^（1/3）， （Y/2）^3=（（1-Y）/4）^2， Y^3/8=（1-2Y+Y^2）/16，2Y^3-Y^2+2Y-1=0， 2Y（Y^2+1）-（Y^2+1）=0， Y=1/2， 2X+3=-1， X=-2

[@tejpalsingh366]

X= -2 only real soln
Rests r complex

[@walterwen2975]

A Nice Exponential Equation: √[(2^(2x + 5)/8] = ³√{[1 - 2^(2x + 3)]/4}, x ϵR
√[(2^(2x + 5)/8] = √[(2²ˣ)(2⁵)/8] = √[(2²)(2ˣ)²] = 2(2ˣ)
[1 - 2^(2x + 3)]/4 = [1 - 8(2ˣ)²]/4 = [2(2ˣ)]³ = 8(2ˣ)³, Let: y = 2ˣ
(1 - 8y²)/4 = 8y³, 32y³ + 8y² - 1 = 0, y³ + (1/4)y² - 1 /32 = 0
(y³ - 1/64) + [(1/4)y² - 1/64] = [y³ - (1/4)³] + (1/4)[y² - (1/4)²] = 0
(y - 1/4)(y² + y/4 + 1/16) + (1/4)(y + 1/4)(y - 1/4) = 0
(y - 1/4)(y² + y/4 + 1/16 + y/4 + 1/16) = (y - 1/4)(y² + y/2 + 1/8) = 0
y = 2ˣ > 0, x ϵR: y² + y/2 + 1/8 > 0; y - 1/4 = 0, y = 2ˣ = 1/4 = 2⁻², x = - 2
Answer check:
x = - 2: 2ˣ = 1/4
√[(2^(2x + 5)/8] = 2(2ˣ), ³√{[1 - 2^(2x + 3)]/4} = ³√{[1 - 8(2ˣ)²]/4}
2(2ˣ) = 2(1/4) = 1/2, ³√[(1 - 1/2)/4] = ³√(1/8) = 1/2; Confirmed
Final answer:
x = - 2

[@RealQinnMalloryu4]

8x^2^10/8=1.2x^1^.2 1^12^1x1^12^1 1^1'x^2^1 x^2^1 (x ➖ 2x+1) (1)^2 ➖ (2)^2{2x+2x ➖ 3+3 ➖ }{ 1 ➖ 4}{4x^2+6} ={3+10x^2}/4 1=13x^2/4=3.1x^2 3^1.1^1x^2^1 3x^2 (x ➖ 3x+2)

[@user-kt1dm9jz5t]

,X=-2

[@user-ee7nw2rx9s]

Самая лучшая замена а=2^(2х+3)
Sqrt (a/2)=sqrt (3)((1-a)/4)
Возводим в 6 степень
а^3/8=(1-а)^2/16
2*а^3-а^2+2*а-1=0
2а(а^2+1)-(а^2+1)=0
2а-1=0, а=2^(-1)=2^(2х+3)
2х+3=-1
х=-2
3 минуты, а не 12 как на видео

[@SidneiMV]

2^(2x) = u
4³u³ = (1 - 8u)²/4²
4⁵u³ = 8²u² - 16u + 1
2¹⁰u³ - 2⁶u² + 2⁴u - 1 = 0
2⁶u²(2⁴u - 1) + (2⁴u - 1) = 0
(2⁴u - 1)(2⁶u² + 1) = 0
2⁴u - 1 = 0 => u = 1/2⁴
2^(2x) = 1/2⁴
2x = -4 => *x = -2*
2⁶u² + 1 = 0 => no real solutions

[@RajeshKumar-wu7ox]

((1+√1-4√2))/4√2),((1-root(1-4√2))/4√2),((-1+root (1+4√2))/4√2),((-1-root(1+4√2))/4√2)