A Nice Exponential Equation Challenge | An Algebra Puzzle!

A Nice Exponential Equation Challenge | An Algebra Puzzle!
Welcome to another exciting algebra challenge! In this video, we'll solve a fascinating exponential equation that will raise your mathematical skills. Can you solve it? Watch the video, try it yourself, and share your solutions in the comments! This is a perfect problem for those preparing for math competitions or anyone who loves a good brain teaser. Don't forget to like, subscribe, and hit the notification bell for more challenging math puzzles!
Topics Covered:
Exponential equations
How to solve exponential equations?
Algebra
Properties of exponents
Algebraic identities
Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Solving quintic equation
Quadratic equation
Discriminant
Real solutions
Additional resources:
• An Interesting Algebra...
• A Nice Algebra Problem...
• Can You Solve This Exp...
• Solving a Tricky Expon...
#matholympiad #exponentialequations #problemsolving #mathcompetition #mathtutorial #mathematics #learnmaths #mathenthusiast #algebra #exponents #algebrachallenge
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Thanks for Watching !!

[@ZhilinChen-my7tp]

X=0, (log3+√5/ log 2)-1, (log3-√5/ log 2)-1

[@RajeshKumar-wu7ox]

X=0,log((3+√5)/2)/log2

[@tejpalsingh366]

x=0; (log(3+-√5)/ log2)-1

[@kassuskassus6263]

Let u=2^x and solve for u. u=1 (double) gives x=0 and u=(3+or-sqrt5)/2 gives x=log((3+or-sqrt5)/log2)-1.

[@user-kp2rd5qv8g]

Let t=2^x. Then, [t^4+8t^2+1]/[[t^3+t] = 5 > t^2[t^2+1/t^2+8]/[t^3+t] = 5 > [(t+1/t)^2+6]/(t+1/t) = 5 > (t+1/t)^2 -5 (t+1/t) + 6 = 0 > t+1/t = 2,3 > x=0 or x = log_2 [(3 +/-sqrt5)/2].

[@user-ny6jf9is3t]

χ=0 διπλη, ή χ=[[log(3+ -ριζα5)]/log2]-1

[@RealQinnMalloryu4]

{4^4x^24^6x^2}={ 8^10x^4+2}= 10^10x^4=/{4^6x^2+4x^2}= 8^6x^4 _10^10x^4/8^6^x^4=1.^2^1^.2^6x^1 1^1.2^1^1^1.2^1^3^2x1^1 11 x3^2 x^3^2.(x ➖ 3x+2 )

[@prateek1.9]

cant you use hindi sir??

[@SidneiMV]

2^x = u
(u⁴ + 8u² + 1)/(u³ + u) = 5
[(u² + 1)² + 6u²]/(u³ + u) = 5
(u² + 1)/u + 6u/(u² + 1) = 5
(u² + 1)/u = w
w + 6/w = 5
w² - 5w + 6 = 0
(w - 2)(w - 3) = 0
w = 2
(u² + 1)/u = 2
u² - 2u + 1 = 0
(u - 1)² = 0 => u = 1
2^x = 1 => *x = 0*
w = 3
(u² + 1)/u = 3
u² - 3u + 1 = 0
u = (3 ± √5)/2
2^x = (3 ± √5)/2
2^(x + 1) = 3 ± √5
(x + 1)ln2 = ln(3 ± √5)
*x = (1/ln2)ln(3 ± √5) - 1*

[@user-nd7th3hy4l]

X=0

[@abcekkdo3749]

X=0,

[@yakupbuyankara5903]

X=0