Solving a Tricky Exponential Equation | Math Olympiad Challenge

Поділитися
Вставка
  • Опубліковано 15 чер 2024
  • Hello My dear family I hope you all are well if you like this video about "Solving a Tricky Exponential Equation | Math Olympiad Challenge" then please do subscribe our channel for more mathematical adventures like this.
    Join us as we tackle a tricky exponential equation from the Math Olympiad! In this video, we break down the problem, explore different approaches, and guide you through the step-by-step solution. Perfect for students preparing for math competitions, educators looking for challenging problems, or math enthusiasts eager to learn more. Don't forget to like, share, and subscribe for more exciting math challenges and solutions!
    Topics covered:
    Exponential equations
    How to solve exponential equations?
    Algebra
    Properties of exponents
    Algebraic identities
    Synthetic division
    Rational root theorem
    Exponential Equation
    Math Olympiad preparation
    Math Olympiad training
    Exponent laws
    Solving cubic equation
    Quadratic equation
    Discriminant
    Complex solutions
    Real solutions
    Timestamps:
    0:00 Introduction
    0:30 Properties of exponents
    2:12 Algebraic manipulations
    4:40 Solving Exponential equation
    5:48 Substitution
    8:23 Solving cubic equation
    11:42 Quadratic equation
    13:15 Solution
    13:25 Verification
    #matholympiad #exponentialequations #mathchallenge #problemsolving #mathcompetition #mathtutorial #olympiadmathematics #mathematics #learnmaths #mathenthusiast
    #algebra
    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 • 6

  • @mohammedsaysrashid3587
    @mohammedsaysrashid3587 25 днів тому +1

    A wonderful introduction.👌.....x=-2 is a real solution

  • @user-kp2rd5qv8g
    @user-kp2rd5qv8g 25 днів тому +6

    Let a = 2^(2x+4). Then, the given equation is written as a^3+2a^2=3. This has a=1 as the only real solution. So, 2x+4=0 > x = -2.

  • @roberttelarket4934
    @roberttelarket4934 25 днів тому +1

    (8^t, 4^t, 2^t) > 0.
    If integers: 1 + 1 + 4 = 6.
    x = -2 does the trick.
    May be other roots.

  • @StaR-uw3dc
    @StaR-uw3dc 25 днів тому +1

    x=-2