A Gigantic Exponential Equation | Can you solve?

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  • Опубликовано: 23 дек 2024

Комментарии • 30

  • @Mehrdad_Basiry
    @Mehrdad_Basiry Год назад +3

    Thanks...
    Please put more of these kinds of questions...🙏🙏🙏.
    I love them alot...🤩🤩🤩.

  • @stevestarcke
    @stevestarcke Год назад +1

    Fun and instructive. Thank you.

  • @sanjaykamath90210
    @sanjaykamath90210 Год назад

    2^2 when Brought one level down 4 and vice versa 2^16386= 2^16384*2^2 =2^16386. Since multiplication is repeated addition. Actually you should use Prime Factors

    • @sanjaykamath90210
      @sanjaykamath90210 Год назад

      The ERROR IS THAT YOU HAVE MULTIPLIED THE RHS BY 2 WITHOUT DIVIDING ON LHS.

  • @mathswan1607
    @mathswan1607 Год назад +2

    x=16

  • @sanjaykamath90210
    @sanjaykamath90210 Год назад

    I think 2173....will chk later🎉

  • @sandrosalustri5677
    @sandrosalustri5677 Год назад +1

    16 exp16 exp16 exp3 (=2 exp3072) is not equal to 2 exp2 exp16386 (=2 exp32772).

    • @pluisjenijn
      @pluisjenijn Год назад +1

      That's because we are missing parentheses in these equations. (2^2)^16 is not 2^(2^16). You are treating the equation as ((x^x)^x)^3 but it's x^(x^(x^3))

    • @sandrosalustri5677
      @sandrosalustri5677 Год назад +1

      Thanks for your time.

  • @math_plus_intelligence
    @math_plus_intelligence Год назад

    good job 👏

  • @notlin1976
    @notlin1976 Год назад

    Fantastic!
    🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷

  • @laithqashoa
    @laithqashoa Год назад

    Nice

  • @sanjaykamath90210
    @sanjaykamath90210 Год назад

    Sorry 2173*2 😮 ie 4346

  • @chmjnationalsuperarmygener8564

    hk bus 16 kwun tong lam tin to mongkok pak king wan

  • @vcvartak7111
    @vcvartak7111 Год назад

    The number is so myriad, it's difficult to verify on calculator

  • @sandrosalustri5677
    @sandrosalustri5677 Год назад

    X=16 is not a solution.

    • @lettucehelper
      @lettucehelper Год назад

      It sure looks like it is.

    • @johntse5770
      @johntse5770 Год назад

      No, 16 is indeed the solution.
      Step 1: Remove the outermost "2^" at the RHS using a substitution
      Let x = 2^n
      (2^n)^{(2^n)^[(2^n)^3]} = 2^(2¹⁶³⁸⁶)
      Known: (a^b)^c = a^(bc)
      (2^n)^{(2^n)^[2^(3n)]} = 2^(2¹⁶³⁸⁶)
      (2^n)^[2^(n · 2^(3n))] = 2^(2¹⁶³⁸⁶)
      2^(n · 2^(n · 2^(3n))) = 2^(2¹⁶³⁸⁶)
      n · 2^(n · 2^(3n)) = 2^(16386)
      Step 2: Repeat Step 1 using a similar substitution
      Let n = 2^t
      2^(t + 2^t · 2^(3 · 2^t)) = 2^(16386)
      t + 2^(t + 3 · 2^t) = 16386
      t + 2^(t + 3 · 2^t) = 2¹⁴ + 2
      Assume t = 2 and t + 3 · 2^t = 14.
      Since 2 + 3 · 2² = 14, t = 2
      -> n = 2² = 4
      -> x = 2⁴ = 16

    • @SyberMath
      @SyberMath  Год назад

      Wow! 😍

    • @pluisjenijn
      @pluisjenijn Год назад

      Please explain yourself instead of making statements

    • @sandrosalustri5677
      @sandrosalustri5677 Год назад

      Thanks for your time.

  • @JXS63J
    @JXS63J Год назад +1

    "So far so good?" you ask. NO! You really need to preedit these before putting on line. Way too many misleading points that should easily be eliminated. Would you have the patience if a student pulled this stuff... for a prepared problem no less?

    • @neuralwarp
      @neuralwarp Год назад +1

      My thoughts exactly. You owe your viewers the courtesy of a properly edited video.

    • @sanjaykamath90210
      @sanjaykamath90210 Год назад

      x^x^x^3
      The prime factors of 16386 are 3*2*2173
      RHS can bbe simplified as :
      LHS 2 ,2173,2,3