Let's Solve A Challenging Exponential Equation for All Reals

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  • Опубліковано 26 чер 2024
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    Let's Solve A Challenging Exponential Equation for All Reals
    In this video, we tackle a challenging exponential equation and solve it for all real numbers. Join us as we explore different techniques and strategies to find the solution. Whether you're a math enthusiast or preparing for exams, this step-by-step tutorial will enhance your problem-solving skills and deepen your understanding of exponential functions. Don't forget to like, subscribe, and hit the bell icon for more math challenges and tutorials!
    Topics covered:
    Exponential equations
    How to solve exponential equations
    Algebra
    Algebraic identities
    Exponential Equation
    Math Olympiad preparation
    Math Olympiad training
    Exponent laws
    Quadratic equation
    Factorization
    Algebraic manipulations
    Real solutions
    Timestamps:
    0:00 Introduction
    1:40 Exponent laws
    2:36 Substitutions
    5:42 Algebraic identities
    9:11 Solutions
    9:25 Verification
    #mathchallenge #exponentialequations #mathtutorial #algebra #problemsolving #mathhelp #learnmath #mathforall #equationsolving #stem #matheducation #studytips #mathtricks #math
    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 !!
    ‪@infyGyan‬

КОМЕНТАРІ • 9

  • @StaR-uw3dc
    @StaR-uw3dc Місяць тому +3

    Nice solution.
    Other approach: AM-GM: (a⁴+b⁴+c⁴+1)/4 ≥ (a⁴∙b⁴∙c⁴∙1)^(¼) = abc. Equality holds only when all four terms are equal ie. a⁴=b⁴=c⁴=1. Hence x=0.

  • @mohammedsaysrashid3587
    @mohammedsaysrashid3587 Місяць тому +2

    A wonderful introduction and explanation.....x=0

  • @alibhukoo5400
    @alibhukoo5400 Місяць тому

    Great method 😊❤❤

  • @qquark99
    @qquark99 Місяць тому +3

    I appreciate the very clever algebraic manipulations. But x = 0 is obvious by inspection. Any expression of similar exponential form in which the numerator and denominator are equal if x = 0 could have been posed as a problem. Proving uniqueness of x = 0 may be a challenge, though.

  • @tunneloflight
    @tunneloflight Місяць тому +1

    X=0. By inspection of constraints, it is the only possible solution.

  • @abcekkdo3749
    @abcekkdo3749 Місяць тому +2

    X=0

  • @sunil.shegaonkar1
    @sunil.shegaonkar1 Місяць тому

    At 6.0 minutes, each individual bracket is equal to 0 is one possibility and that leads to the solution, x = 0.
    But there are three other possibilities that I will try to solve, when one bracket is 0, other two brackets can have opposite sign.

    • @ramank.g.s4209
      @ramank.g.s4209 Місяць тому

      But square cannot be negative hence all equal to 0

  • @alibhukoo5400
    @alibhukoo5400 Місяць тому +1

    x=0 is a solution 😊❤