An Interesting Exponential Equation

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  • Опубликовано: 17 янв 2025

Комментарии • 21

  • @山山-y4q
    @山山-y4q Месяц назад +3

    If you align the left and right expressions by aligning the shape of the binomial product by 2,3 and pay attention to the exponent, you can find that the exponent of the left side 2 is x = 1.
    Now x = 1 is
    It is certain that it is an answer. Even so, if you check if the index of 3 is 1, it will be an answer.

  • @scottleung9587
    @scottleung9587 Месяц назад +1

    Nice!

  • @chillwhale07
    @chillwhale07 Месяц назад +3

    Second solution is approximately -5.17 😊

    • @Qermaq
      @Qermaq Месяц назад +1

      It's frighteningly close to -206797/40000.

    • @rakenzarnsworld2
      @rakenzarnsworld2 Месяц назад +1

      ​@@QermaqLol I thought it was irrational number

    • @chillwhale07
      @chillwhale07 Месяц назад +1

      @Qermaq wow that's more accurate.

    • @Qermaq
      @Qermaq Месяц назад

      @@rakenzarnsworld2 It is. It's that fraction plus something like 0,00005 something.

    • @aaalekseev
      @aaalekseev Месяц назад

      @@rakenzarnsworld2it is

  • @AlexIohannsen
    @AlexIohannsen Месяц назад +1

    x ≠ -2;
    2^(x - 1) * 3^(3x/(x + 2) - 1) = 1
    take log_2 of both sides:
    (x - 1) + (2x - 2) * log_2(3)/(x + 2) = 0
    multiply both sides by (x + 2) ≠ 0
    (x - 1) (x + 2 + 2log_2(3)) = 0
    x = 1 or
    x = - 2 - 2log_2(3) = - log_2(36)

  • @Quest3669
    @Quest3669 Месяц назад +4

    X= 1

  • @TejasDhuri-p8z
    @TejasDhuri-p8z Месяц назад +2

    X=1 by looking at it

    • @frendlyleaf6187
      @frendlyleaf6187 Месяц назад +3

      Ah yes any engineer's favourite method: solution by observation

    • @chillwhale07
      @chillwhale07 Месяц назад +3

      You can't solve by looking at it bro 😂

  • @Qermaq
    @Qermaq Месяц назад

    Good one!
    Hey, off-topic, but wanted to ask: I've been playing with cubics where the a term is 1 and the c term is 0, so x^3 + bx^2 + d = 0. I "discovered" (I'm sure I am not the first) a parameterization where if the roots are ak, a(k^2 - k) and a(1-k), the resulting cubic is x^3 - a(k^2 - k + 1)x^2 + (a^3)(k^2)((k - 1)^2) = 0. So if a and k are 2, we get x^3 - 14x^2 + 288 = 0, and the roots are 6, 12 and -4. Have you seen something like this? Is this going to generate all these depressed cubics, or just some? Is this useful to leverage to solve these? Anyway, I found it interesting and fun.

  • @에스피-z2g
    @에스피-z2g Месяц назад

    Solution by insight
    2×3=6
    x=1

  • @dbaznr
    @dbaznr Месяц назад

    Just by sight, x = 1

  • @giuseppemalaguti435
    @giuseppemalaguti435 Месяц назад

    x=1,x=-ln36/ln3...no,devo rifare i calcoli..x=-ln36/ln2=-5,17..ecco,this Is correct

  • @rakenzarnsworld2
    @rakenzarnsworld2 Месяц назад

    x = 1

  • @nasrullahhusnan2289
    @nasrullahhusnan2289 Месяц назад

    2^x•27^x/(x+2)] 🎉😂[2^(x-1)][3^{3x/(x+2)}-1]=1
    2^(x-1)=1 --> x=1
    3^[{3x/(x+2)}-1]=1
    [3x/(x+2)]-1=0 --> 3x=x+2
    x=1