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A Nice Exponential Equation | Math Olympiad Training
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Ай бұрынA Nice Exponential Equation | Math Olympiad Training
Welcome to another intriguing math challenge! In this video, we dive into a beautiful exponential equation from the Math Olympiad Prep. This problem will test your understanding of exponential functions and your problem-solving abilities.
Are you ready to take on this challenge? Watch the video, attempt to solve the equation on your own, and then compare your solution with ours. Don’t forget to share your approach and solution in the comments below!
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Topics covered:
Exponential equations
How to solve exponential equations?
Algebra
Exponents
Factorization
Properties of exponents
Algebraic identities
Radicals
Cubic equations
Math Olympiad preparation
Math Olympiad training
Exponent laws
Real solutions
Additional Resources:
• A Nice Exponential Equ...
• An Interesting Exponen...
• An Interesting Factori...
• Can You Solve This Exp...
#exponentialequations #cubicequation #mathematics #math #matholympiad #problemsolving #exponents #radical #algebra
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Thanks for Watching !!
@mohammedsaysrashid3587 Ай бұрын
It was a wonderful introduction and clearly explaining....thanks, Sir 🙏....x=4
@user-kp2rd5qv8g Ай бұрын
Let 2^(x/2) = t. Then, the given equation is equivalent to t^3-2t^2-32=0 which has t=4 as the only real solution. Thus, x/2 = 2 > x=4.
@samah9496 Ай бұрын
2^(3x/2)-2×2^x-32=0 X=2^(x/2) X³-2X²-32=0 Pour X=4 4³-2×4²-32=0 (X-4)(X²+2X+8)=0 X²+2X+8=0 solution complexe S={4} X=2^(x/2)=4=2² x/2=2 x=4
@user-ny6jf9is3t Ай бұрын
χ=4
@boy-bf3si Ай бұрын
X=4 is answer
@user-kt1dm9jz5t Ай бұрын
x=4
@tejpalsingh366 Ай бұрын
X= 4 only real soln Rest r complex
@walterwen2975 Ай бұрын
A Nice Exponential Equation: (√2)ˣ - 2⁵⁻ˣ = 2, x ϵR; x = ? (√2)ˣ > 0, No complex and/or imaginary value root (√2)ˣ - 2⁵⁻ˣ = (√2)ˣ - 2⁵/2ˣ = (√2)ˣ - 2⁵/(√2ˣ)² = 2, [(√2)ˣ]³ - 2(√2ˣ)² - 2⁵ = 0 Let: y = (√2)ˣ, [(√2)ˣ]³ - 2(√2ˣ)² - 2⁵ = y³ - 2y² - 32 = 0; y = (√2)ˣ > 0, y ϵR (y³ - 4y²) + 2(y² - 16) = y²(y - 4) + 2(y + 4)(y - 4) = (y - 4)(y² + 2y + 8) = 0 y² + 2y + 8 > 0; y - 4 = 0, y = 4, (√2)ˣ = 2ˣ⸍² = 4 = 2², x/2 = 2; x = 4 Answer check: (√2)ˣ - 2⁵⁻ˣ = 2⁴⸍² - 2⁵⁻⁴ = 4 - 2 = 2; Confirmed Final answer: x = 4
@RealQinnMalloryu4 Ай бұрын
(2x)^2=2x^2 ➖ (2 x)^5= 32x^5 {4x^2 ➖ 32x^5}= 28x^3 3^8x^3 1^2^3x^3 2^3^1x^3^1 21^1x^3^1 2x^3 (x ➖ 3x+2)
@RajeshKumar-wu7ox Ай бұрын
X=4
@RajeshKumar-wu7ox Ай бұрын
And another solution is comlex
@kassuskassus6263 Ай бұрын
x=4
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