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A Nice Exponential Equation | An Algebra Challenge
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Ай бұрынA Nice Exponential Equation | An Algebra Challenge
Welcome to infyGyan !! In this video, we'll be solving a nice exponential equation that's both challenging and fun. Perfect for algebra enthusiasts and students preparing for advanced math competitions, this tutorial will take you step-by-step through the process, helping you understand the concepts and techniques needed to tackle exponential equations with confidence.
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
Properties of exponents
Properties of surds
Algebraic identities
Radicals
Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Real solutions
#educational #exponentialequations #olympiadmath #mathematics #math #matholympiad #problemsolving #radical #algebra
Additional resources:
• A Fascinating Radical ...
• Solving a Tricky Expon...
• Let's Solve A Challeng...
• Tough Exponential Equa...
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.
Thanks for Watching !!
@mohammedsaysrashid3587 Ай бұрын
It was wonderful explanation thanks for sharing Sir 🙏....x=2/3
@user-kp2rd5qv8g Ай бұрын
The equation can be written as 2^(3x) + 4 = 1/8 2^9x. Let t = 2^3x > t^3= 8t+32 > t=4 is a solution. [t^3-8t-32]/(t-4) = t^2+4t+8. t^2+4t+8 = 0 has the solutions t = -2 +/- 2 i which are complex and not acceptable. Thus t=4 > x = 2/3.
@dorkmania Ай бұрын
Simplifying RHS to (2^3x)/2 and substituting for LHS radical with t³ = (2^3x) + 4, the equation becomes (t³)^⅓ = (t³ - 4)/2 => 2t = t³ - 4 => t³ - 2t - 4 = 0 Using RRT and SDM, we get t = 2 as the only real solution. 2^3x - 4 = (2)³ => x = ⅔
@woobjun2582 Ай бұрын
The given can be ³√(t +4) = t/2 by letting t = 2³ˣ Cubing and rearringing, t³-8t -32 = 0; t³ -64 -8t +32 = 0; (t³ -4³) -8(t -4) = 0; (t -4)(t² +4t +16) -8(t -4) =0; (t -4)(t² +4t +8) =0 That is, t -4 =0 ; t = 4 or t² +4t +8 = 0; t = cmplx Therefore, t = 2³ˣ = 4 = 2² Finally 3x =2; x = 2/3
@yakupbuyankara5903 Ай бұрын
2/3
@tejpalsingh366 Ай бұрын
X=2/3 only real soln Rest is complex
@RajeshKumar-wu7ox Ай бұрын
X=2/3 rest solution find out for dificult.
@abcekkdo3749 Ай бұрын
X=2/3
@walterwen2975 Ай бұрын
An Algebra Challenge: ³√(8ˣ + 4) = √4³ˣ⁻¹, x ϵR; x = ? ³√(8ˣ + 4) = ³√(2³ˣ + 4) = √4³ˣ⁻¹ = (√4³ˣ)/2, 2[³√(2³ˣ + 4)] = √4³ˣ = √2⁶ˣ Let: y = 2³ˣ; 2[³√(y + 4)] = √(y²) = y, y³ = 8(y + 4), y³ - 8y - 32 = 0; y > 0 y³ - 64 - 8y + 32 = (y³ - 4³) - (8y - 32) = (y - 4)(y²+ 4y + 16) - 8(y - 4) = 0 (y - 4)(y²+ 4y + 16 - 8) = (y - 4)(y²+ 4y + 8) = 0, y = 2³ˣ, x ϵR; y²+ 4y + 8 > 0 y - 4 = 0, y = 4 = 2³ˣ, 2³ˣ = 4 = 2², 3x = 2; x = 2/3 Answer check: ³√(8²⸍3 + 4) = ³√8 = 2, √4³ˣ⁻¹ = √4³ˣ⁻¹ = √4 = 2; Confirmed Final answer: x = 2/3
@RealQinnMalloryu4 Ай бұрын
{16^8+12 }= 28^8 7^4^2^3 7^12^2^2^3^1 1^11^12^1^1 2^1 (x ➖ 2x+1){ 64^27 ➖ 1} =64^26 8^8^2^13 2^32^3^2^13^1 1^11^12^1^1 2^1 (x ➖ 2x+1)
@kassuskassus6263 Ай бұрын
One real solution x=2/3.
@mulla_modi Ай бұрын
X=2/3