Canada | A Nice Algebra Problem | Math Olympiad

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  • Опубликовано: 1 дек 2024

Комментарии • 15

  • @ChavoMysterio
    @ChavoMysterio Месяц назад +2

    9^(n+1)-9^(n-1)=20
    9(9ⁿ)-[(9ⁿ)/9]=20
    81(9ⁿ)-9ⁿ=180
    80(9ⁿ)=180
    9ⁿ=2.25
    3²ⁿ=1.5²
    2n=log_3(1.5²)
    2n=2[log_3(1.5)]
    n=log_3(1.5) ❤

    • @SALogics
      @SALogics  Месяц назад +1

      Very nice! ❤

  • @에스피-z2g
    @에스피-z2g Месяц назад +1

    9^(x-1)×(81-1)=20
    9^(x-1)=1/4
    9^x=9/4
    x=log(9/4)/log9
    =(log9-log4)/log9
    =1-log4/log9
    =1-log2/log3

    • @SALogics
      @SALogics  Месяц назад +1

      Very nice! ❤

  • @matheducare
    @matheducare Месяц назад +1

    Nice math.
    It improve skill.

    • @SALogics
      @SALogics  Месяц назад +1

      Very nice! ❤

  • @dblogp
    @dblogp Месяц назад +1

    Greetings, excellent videos, but I have a question: What application do you use for classes?

  • @stpat7614
    @stpat7614 19 дней назад

    9^(x + 1) - 9^(x - 1) = 20
    9^x*9^1 - 9^x*9^(-1) = 20
    9^x*9^1 - 9^x/9 = 20
    9^x*9*9 - 9^x/9*9 = 20*9
    9^x*81- 9^x = 180
    9^x(81 - 1) = 180
    9^x*80 = 180
    9^x = 180/80
    9^x = 9/4
    x = log9(9/4)
    x = log9([3/2]^2)
    x = 2*log9(3/2)

  • @trojanleo123
    @trojanleo123 Месяц назад +1

    x = 1 - ln2/ln3

    • @SALogics
      @SALogics  Месяц назад +1

      Very nice! ❤

  • @jamesharmon4994
    @jamesharmon4994 Месяц назад +1

    By observation, x is greater than 0 and less than 1.

    • @SALogics
      @SALogics  Месяц назад +2

      Very nice! ❤

    • @jamesharmon4994
      @jamesharmon4994 Месяц назад +1

      @@SALogics Upon further observation, between 0 and 1/2. At one half, it becomes 27 - (1/3), which is greater than 20.