Brazil | A Nice Algebra Problem | Math Olympiad

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  • Опубликовано: 24 окт 2024

Комментарии • 6

  • @mohamadpakzamir
    @mohamadpakzamir 2 дня назад +1

    64---25=39
    ((2)^3/2)^)^x^1/2=2^6
    X=16
    (5^)^y^1/2=25=5^2
    y=16

  • @Quest3669
    @Quest3669 2 дня назад +2

    2^6-5^2= 39 gives x= y= 16: x>0

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 2 дня назад +1

    (8)^2 ➖ (5)^2={64 ➖ 25}=39 3^13 3^13^1 3^1^1 3^1 ( xy ➖ 3xy+1).

  • @gaiatetuya92
    @gaiatetuya92 2 дня назад

    (x,y)=( (16/9)(log(2)20)^2,16(log(5)19)^2 )

  • @daakudaddy5453
    @daakudaddy5453 2 дня назад +1

    Please mention in the beginning that x and y are integers.
    To solve for 2 variables, you need 2 equations, or rather 2 independent pieces of information. The fact that x and y are both integers is that critical missing piece of information that you casually assume in the middle of the solution.

    • @vijaymaths5483
      @vijaymaths5483  День назад

      Ok 👍 Thank you for your valuable feedback