Japanese | A Nice Algebra Problem | Math Olympiad

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  • Опубліковано 12 гру 2024

КОМЕНТАРІ • 9

  • @narenchennai6875
    @narenchennai6875 6 днів тому +3

    (1/2)^x = x^(1/2)
    taking log, (x)log(1/2) = (1/2)log(x)
    => log(1/2)/(1/2) = log(x)/(x)
    => x=1/2

    • @tander6435
      @tander6435 6 днів тому +1

      Just compare the base and the power

    • @SALogics
      @SALogics  День тому +1

      Very nice! ❤

  • @에스피-z2g
    @에스피-z2g 6 днів тому +1

    (1/2)^x=x^(1/2)
    (1/4)^x=x
    1/4=x^(1/x)
    x=1/2

  • @narsinhapotdar7215
    @narsinhapotdar7215 6 днів тому +1

    Very good

  • @walterwen2975
    @walterwen2975 6 днів тому +1

    Japanese, Math Olympiad: (1/2)ˣ = x¹⸍²; x =?
    x > 0, Let: a = 1/2
    (1/2)ˣ = aˣ = x¹⸍² = xᵃ, xᵃ = aˣ, (xᵃ)¹⸍⁽ᵃˣ⁾ = (aˣ)¹⸍⁽ᵃˣ⁾, x¹⸍ˣ = a¹⸍ᵃ; x = a = 1/2
    Answer check:
    x = 1/2: (1/2)ˣ = (1/2)¹⸍² = x¹⸍²; Confirmed
    Final answer:
    x = 1/2