Math Olympiad | A Nice Algebra Problem | A Nice Exponential Equation

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  • Опубликовано: 11 ноя 2024

Комментарии • 16

  • @sy8146
    @sy8146 Месяц назад +4

    Thank you for explaining. x=66, y=1 is another solution.
    If 33^(1/16) is admitted, irrational numbers are OK. If so, x=√69, y=4 can be one of the solutions.

    • @SALogics
      @SALogics  Месяц назад +1

      You are welcome! ❤

  • @YAWTon
    @YAWTon Месяц назад +1

    There are as many solutions as there are positive real numbers. If y>0 then
    (x,y)=(exp([y+log(65)]/√y,) y) is a solution.
    There are integer solutions, which are not shown in the clip: (-3, 16), (66,1).
    One of the integer solutions gets lost around @7:16, when the author concludes x=3 from x^2=3^2. This is a popular mistake.

    • @SALogics
      @SALogics  Месяц назад +1

      You are right! ❤

  • @rousseau049
    @rousseau049 Месяц назад +1

    X=66 and Y=1 also balances the equation and hence a valid solution...why these numbers never figured in your solution? Any reason?

    • @SALogics
      @SALogics  Месяц назад

      You are right! I forgot to include that in solution ❤

  • @davidpatterson5426
    @davidpatterson5426 Месяц назад +1

    x = 3 & y = 16

    • @SALogics
      @SALogics  Месяц назад +1

      You are right! ❤

  • @АндрейПергаев-з4н
    @АндрейПергаев-з4н Месяц назад +3

    Ответ не верный
    По крайней мере х=66, у=1 тоже решение, а его нет как решения.

    • @SALogics
      @SALogics  Месяц назад +1

      Да, ты прав! ❤

  • @stas1ism
    @stas1ism Месяц назад +1

    Obviously, wrong decision

    • @SALogics
      @SALogics  Месяц назад +1

      Please identify the reason. ❤

    • @mauriziograndi1750
      @mauriziograndi1750 Месяц назад +1

      I appreciate all your calculations but it was obvious at a glance that x=3, y=16, nevertheless we appreciate your method.

    • @SALogics
      @SALogics  Месяц назад +1

      @@mauriziograndi1750 Thanks a lot ❤❤

  • @9허공
    @9허공 Месяц назад +1

    You first define the domain of x and y. it should be positive integers.
    at 2:42 I understand b = √y is integer, but how can you guarantee a = x^(√y/2) is integer?
    if x is not power of 2 and √y is odd , a is irrational.
    in that case there are infinitely many combinations of (a+b)x(a-b).

    • @SALogics
      @SALogics  Месяц назад +1

      Thanks for your feedback! ❤