Solving A Homemade Exponential Equation

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  • เผยแพร่เมื่อ 17 ก.ย. 2024
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ความคิดเห็น • 15

  • @vladimirkaplun5774
    @vladimirkaplun5774 16 วันที่ผ่านมา +3

    I do not see what you see. if c1 there is only one

  • @briancatanzaro6053
    @briancatanzaro6053 16 วันที่ผ่านมา +2

    I am happy with keeping it "real".

  • @seanfraser3125
    @seanfraser3125 16 วันที่ผ่านมา +1

    Raise both sides to the 5th power:
    x^(15x^15) = 2^(5*32) = 2^(5*2^5)
    Using exponent rules, we can rewrite this equation like this:
    (x^15)^(x^15) = (2^5)^(2^5)
    The function f(u) = u^u is one-to-one for u > 1, so the only solution to u^u = 32^32 is the obvious one, u=32. Thus
    x^15 = 2^5.
    Taking the 15th root of both sides gives us x = 2^(1/3).

  • @walterwen2975
    @walterwen2975 16 วันที่ผ่านมา +2

    Solving A Homemade Exponential Equation: x^(3x¹⁵) = 2³²; x =?
    [x^(3x¹⁵)]⁵ = (2³²)⁵ = (2^2⁵)⁵, x^(15x¹⁵) = 2^[5(2⁵)]
    2^[5(2⁵)] = 2^[15(2¹⁵⸍³)/3] = (2¹⸍³)^[15(2¹⸍³)¹⁵] = (³√2)^[15(³√2)¹⁵]
    x^(15x¹⁵) = (³√2)^[15(³√2)¹⁵]; x = ³√2
    Answer check:
    x = ³√2: x^(3x¹⁵) = (³√2)^[3(³√2)¹⁵] = [(³√2)³]^[(³√2)³]⁵ = 2^2⁵ = 2³²; Confirmed
    Final answer:
    x = ³√2

  • @Psykolord1989
    @Psykolord1989 6 วันที่ผ่านมา

    Before watching:
    Looking at the problem before actually going into calculations, my first thought was 2^(1/3) power, i.e. cube root of 2. Why? Because of the exponents on the left.
    3(x^15) struck me as awfully specific, and 2^5 = 32. So, I started with this idea and went from there.
    Plugging in X=2^(1/3) gives us x^15 = (2^(1/3))^15. Recalling that (a^m)^n = a^(mn), we can simplify to 2^(15/3) = 2^5 = 32
    This is the exponent we want. We now have (2^(1/3))^(3*32) = 2^(32*3/3) = 2^32.
    This is admittedly doing the problem quite literally backwards. This strategy will very likely not work for most people; it just happened to work for me this time. (And this is of course not counting any complex solutions).
    Still, at the end of the day, we arrive at x = 2^(1/3)

  • @vighnesh153
    @vighnesh153 15 วันที่ผ่านมา

    x^x = y^y => x = y is one of the solutions. There can be other solutions as well. For example. x = 1/2 and y = 1/4 are also possible solutions where x != y. How to find other solutions?

  • @rakenzarnsworld2
    @rakenzarnsworld2 16 วันที่ผ่านมา

    x = 2^(1/3)

  • @scottleung9587
    @scottleung9587 16 วันที่ผ่านมา

    Got it!

  • @giuseppemalaguti435
    @giuseppemalaguti435 16 วันที่ผ่านมา

    x=e^(W(160ln2)/15)=2^(1/3)

  • @yoav613
    @yoav613 16 วันที่ผ่านมา

    Nice! Time is runing and time is money😂💯

    • @SyberMath
      @SyberMath  14 วันที่ผ่านมา

      Sure is!

  • @phill3986
    @phill3986 16 วันที่ผ่านมา

    👍🔥😁✌️👏👏✌️😁🔥👍

  • @SidneiMV
    @SidneiMV 16 วันที่ผ่านมา +1

    x³^x¹⁵ = 2^2⁵
    x³ = 2 => *x = ∛2*
    but if you don't believe it..
    x³^x¹⁵ = 2^2⁵
    x¹⁵lnx³ = 2⁵ln2
    x¹⁵lnx¹⁵ = 2⁵ln2⁵
    W(x¹⁵lnx¹⁵) = W(2⁵ln2⁵)
    lnx¹⁵ = ln2⁵
    x¹⁵ = 2⁵ => x = ∛2

  • @AshuTosh-mx1gr
    @AshuTosh-mx1gr 16 วันที่ผ่านมา

    145th viewer 😅

    • @SyberMath
      @SyberMath  14 วันที่ผ่านมา

      Wow! 😁