A Nice Olympiad Math Algebra Problem | A Nice Exponential Equation

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  • Опубликовано: 24 дек 2024

Комментарии • 10

  • @davidseed2939
    @davidseed2939 День назад +2

    2^(xx)=2x square both sides
    2^(2xx) =4xx subs a=2xx
    2^a=2a
    take log base 2
    a =1+log_2(a)
    a=1 or 2
    x=1/√2 or 1

  • @daniellerosalie2155
    @daniellerosalie2155 13 часов назад +1

    I solved 80 percent on my own. Merry Christmas, from a math minor an

    s psychology over doing it

    • @SALogics
      @SALogics  3 часа назад +1

      Very nice! ❤

  • @DJ.Nihad6174
    @DJ.Nihad6174 День назад +4

    x =1 is not unique. you lost one solution. you lost one case.
    2^(x² -1) =x
    2^x² /2 =x
    1 /2 =x×2^-x²
    squaring both sides.
    1 /4 =x²×2^-2x²
    multiplying by -2 both sides.
    -1 /2 =-2x²×2^-2x²
    -1 /2 =-2x²×(e^ln2)^-2x²
    -1 /2 =-2x²×e^(-ln2×2x²)
    -ln2 /2 =-2x²ln2×e^(-ln2×2x²)
    ln(1 /2) /1 /2 =-2x²ln2×e^(-ln2×2x²).
    apply both sides lambert W function.
    ln(1 /2) =-2x²ln2
    -ln2 =-2x²ln2
    2x² =1, x =1 /√2.
    we reject negative solution.
    verify this solution.
    2^(1 /√2)² -1 =1 /√2
    2^(1 /2 -1) =1 /√2
    2^(-1 /2) =1 /√2.
    this is second solution.
    x1 =1 /√2.
    x2 =1.

    • @bagumnazmapervin1663
      @bagumnazmapervin1663 День назад +1

      Yes correct value of x is one by root two

    • @davidseed2939
      @davidseed2939 День назад +1

      i dont think you need Lambert function.
      starting from -1/2=-2x^2 . 2^(2x^2)
      1/2=a.2^- a , where a=2x^2
      rearranging
      2^a=2a
      by inspection a=1 and a=2 are solutions.
      this corresponds to x=1/√2 and x=1

    • @SALogics
      @SALogics  День назад +1

      Very nice! ❤

  • @bagumnazmapervin1663
    @bagumnazmapervin1663 День назад +1

    I think one by root two is also a correct value of x in this equation please justify it

    • @SALogics
      @SALogics  День назад +1

      Yes, you are right!❤