A Nice Exponential Equation | Thailand Junior Math Olympiad

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Күн бұрын

A Nice Exponential Equation | Thailand Junior Math Olympiad
Welcome to another exciting math challenge! In this video, we dive into "A Nice Exponential Equation," perfect for those preparing for Math Olympiads or who loves solving algebra challenge. We'll walk through the problem step-by-step, breaking down the complexities of exponential equations and providing you with tips and strategies to solve similar problems on your own. Whether you're a student or math enthusiast, this video is sure to sharpen your algebra skills and test your problem-solving abilities.
Join us as we explore this intriguing problem and enhance your algebra skills. If you find the video helpful, don't forget to like, subscribe, and hit the bell icon for more exciting math challenges and tutorials.
Topics Covered:
Exponential equations
How to solve exponential equations?
Algebra
Factorization
Properties of exponents
Algebraic identities
Radicals
Cubic equation
Exponential Equation
Math Olympiad
Math Olympiad preparation
Math Olympiad training
Real solutions
#educational #exponentialequations #olympiadmath #mathematics #math #matholympiad #problemsolving #radical #algebra #olympics
Additional Resources:
• A Nice Exponential Equ...
• A Wonderful Factorial ...
• A Nice Simplification ...
• Tough Exponential Equa...
Time-Stamps:
00:00 Introduction
00:44 Exponent laws
04:48 Substitution
05:30 Cubic equations
06:02 Factorization
08:10 Quadratic equation
09:30 Discriminant
10:25 Real solutions
11:10 Verification
Don't forget to like this video if you found it helpful, subscribe to our channel for more Olympiad-focused content, and ring the bell to stay updated on our latest math-solving sessions.
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Пікірлер: 13

@RyanLewis-Johnson-wq6xs 27 күн бұрын

I did it in my head.

@kassuskassus6263 27 күн бұрын

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.

@RyanLewis-Johnson-wq6xs 27 күн бұрын

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

@user-kp2rd5qv8g 27 күн бұрын

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.

@user-ny6jf9is3t 27 күн бұрын

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

@tejpalsingh366 27 күн бұрын

X= -2 only real soln Rests r complex

@walterwen2975 26 күн бұрын

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 27 күн бұрын

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 27 күн бұрын

,X=-2

@davidshen5916 15 күн бұрын

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

@user-ee7nw2rx9s 26 күн бұрын

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

@RajeshKumar-wu7ox 27 күн бұрын

((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)

@SidneiMV 25 күн бұрын

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