A Very Nice Math Olympiad Problem | Solve for x? | Algebra Equation

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  • Опубліковано 12 січ 2025

КОМЕНТАРІ • 2

  • @zihaoooi787
    @zihaoooi787 17 днів тому

    attempting this before the video starts
    8^log(x) - 2^log(x) = 5!
    5! is trivially 120.
    8^log(x) - 2^log(x) = 120
    (2^3)^log(x) - 2^log(x) = 120
    (2^log(x))^3 - 2^log(x) = 120
    let u = 2^log(x)
    u^3 - u = 120
    u(u^2 - 1) = 120
    (u-1)u(u+1) = 120
    after factorizing 120 and using the simple process of using your eyes, we get 4 * 5 * 6 = 120.
    therefore u = 5
    2^log(x) = 5
    log(x) = log_2(5)
    x = 10^log_2(5) approx. 209.85924

  • @pspprabhat
    @pspprabhat 9 днів тому

    =>(2^log×)^3-2^log×
    =120=5^3-5
    =>2^logx=5
    &logx=log@2 (5)
    =>×=10^log@2(5)