A Nice Exponential Equation Challenge | An Algebra Puzzle!

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  • Опубліковано 4 лип 2024
  • 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
    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 !!

КОМЕНТАРІ • 12

  • @ZhilinChen-my7tp

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

  • @kassuskassus6263
    @kassuskassus6263 28 днів тому +1

    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.

  • @tejpalsingh366

    x=0; (log(3+-√5)/ 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].

  • @RajeshKumar-wu7ox

    X=0,log((3+√5)/2)/log2

  • @user-ny6jf9is3t
    @user-ny6jf9is3t 28 днів тому +1

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

  • @prateek1.9
    @prateek1.9 28 днів тому +1

    cant you use hindi sir??

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 28 днів тому

    {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 )

  • @SidneiMV
    @SidneiMV 28 днів тому +1

    2^x = u

  • @user-nd7th3hy4l
    @user-nd7th3hy4l 28 днів тому

    X=0

  • @yakupbuyankara5903
    @yakupbuyankara5903 28 днів тому

    X=0