@@user-sr5lw3bv9приехали. Электромагнетизм чуть менее, чем полностью на комплексных числах расписывается. ОТО в 4-мерном пространстве на обощении комплексных - кватернионах - строится
That's very good, although there's a small error at the very end. The final step should use an approximately equal sign, not an equal sign. 1.585 is an approximation.
(2^a-3)*(1+(2^a-3)+(2^a-3)*(2^a-3))=0 (2^a-3)=0 a=log(2)3=1,585 Зачем нужно было решение этой элементарнейшей задачи растягивать на 10 минут? Красота, кроется в простоте, а то что показали это не красиво.
If you were going to reject the complex roots, why didn't you stop at 6:15 as soon as you saw that 16-52 (under the square root symbol) would be negative? Why go on for another minute and a half to calculate what those rejected complex roots would be?
почему отказались от комплексного корня? Если при решении мы не отказались сразу от комплексных числе, то должны были воспользоваться понятием вектора. А вектор существует на плоскости, стало быть и комплексное решение этого уравнение СУЩЕСТВУЕТ!
@@user-sr5lw3bv9 Дружище, если в твоей практике не применяются, то это не значит, что у всех. У инженеров в расчетах электрики, в гидравлике, аэродинамике ты шагу не ступишь без них. Другое дело, что в ролике школьная задача, а в школе мнимые числа не используют
@@user-sr5lw3bv9На самом деле комплексные числа применяются во многих областях, например, в электротехнике для описания колебательных процессов. Просто автору ролика надо было либо указать, что при D
This is nit picking. But that is what math is all about. So, I definitely agree. As stated, there are three solutions to the problem. And the answer given is not correct. It is an approximation, to three decimal positions (or four characters, if you wish). I believe that the correct answer should have stopped at 'a equal log of three to the base 2'. Thank you and yours for your efforts. May you and yours stay well and prosper.
After t = 2^a, we have t+t²+t³=39. By guessing (factor of 39 is +-1, +-3, +-13), we get t=3 Then we divide t³+t²+t-39 by (t-3) to get t²+4t+13 and since D=4²-4*13
If only real solutions for t are wanted then really it's overkill to find the complex ones. From the discriminant, or even finding the turning point, it is clear this quadratic has no real roots.
We don't want to use any formula, See this easy method, 2^a+4^a+8^a=39 2^a+(2^2)^a+(2^3)^a=39 2^a+2^2a+2^3a=39 Here 2^a is taken a common, 2^a(1+2^2+2^3)=39 Adding this, 2^a(1+4+8)=39 2^a(13)=39 2^a=13 Taking log on both side log 2^a=log 3 a log 2= log 3 So, a= log3/log 2 a= log base(2) 3 is the answer
I believe you have a mistake when you factored the 2^a out . When you factor 2^a out of 2^a + 2^(2a) + 2^(3a) the result is not 2^a(1 + 2^2 + 2^3) or 2^a(1 + 4 + 8). The result is 2^a(1 + 2^a + 2^(2a)). If your statement were true, then let's multiply your statement back and see if we can recover the original statement . . . 2^a(1 + 4 + 8) = 2^a(1 + 2^2 + 2^3) = 2^a + 2^a(2^2) + 2^a(2^3) = 2^a + 2^(a+2) + 2^(a+3)), and clearly that is not the same as 2^a + 2^(2a) + 2^(3a). Remember when you multiply like bases you ADD exponents!
Considerando que (a^n)•(a^m)=a^(n+m), entonces si factorizas 2^a dentro del paréntesis te queda (1+(2^a)+(2^a)^2)), si pensaste que después de factorizar te quedaba (1+2^2+2^3) pues tuviste un error de conceptos ahí y llegaste solo por casualidad al resultado
2^a + 4^a + 8^a = 39 so 2^a + 2^2a + 2^3a = 39 Let x = 2^a then x"3 + x^2 + x - 39 = 0 but 27 + 9 + 3 = 3 and x^3 - 3x^2 + 4x^2 - 12x + 13x - 39 =0 so x =3 or x^2 + 4x + 13 = 0 in which case x = (-4+-sqrt(16-52))/2 = -2+-6i
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
What a Perfect solution it is!
Complex solutions shouldn't be rejected, unless it's specified, they're mathematically valid
В них нет никакого практического смысла
@@user-sr5lw3bv9приехали. Электромагнетизм чуть менее, чем полностью на комплексных числах расписывается. ОТО в 4-мерном пространстве на обощении комплексных - кватернионах - строится
Would it just be easier just to write log(base 2) of 3?
From t³ + t² + t - 39 = 0, the rational root theorem will yield t = 3. Easy from there.
Yes, agreed. I tried to find a back door by using the geometric sum formula, but it led me back to the same cubic polynomial.
That's very good, although there's a small error at the very end.
The final step should use an approximately equal sign, not an equal sign. 1.585 is an approximation.
Ingenious! Simply ingenious👍👍
最後にLogを近似値の小数で表すのには納得いかない
There is no way to express this number as a rational fraction in any number system.
(2^a-3)*(1+(2^a-3)+(2^a-3)*(2^a-3))=0
(2^a-3)=0
a=log(2)3=1,585
Зачем нужно было решение этой элементарнейшей задачи растягивать на 10 минут? Красота, кроется в простоте, а то что показали это не красиво.
If you were going to reject the complex roots, why didn't you stop at 6:15 as soon as you saw that 16-52 (under the square root symbol) would be negative? Why go on for another minute and a half to calculate what those rejected complex roots would be?
2^a=u (u>0, a≠0)
→ u+u²+u³=39
u(1+u+u²)=3·13
∴ u=3, 1+u+u²=13
2^a=3 → a=log₂3
It’s correct answer!
Beautiful music!
No reason given for rejecting complex solutions
👍
Es muchísimo más simple de resolver.
Al inicio, factorizando por 2^a queda
2^a * (1 + 2^2 + 2^3) = 39
2^a * 13 = 39
2^a = 3
a = log 3 / log 2
that´'s all folks
@@YAWTon You just demonstrated that 1 is not a solution. It does not negate the previous solution.
Much easier and easier to understand. I solved it the same way in under 2 minutes.
Exactly 😅😂
@@YAWTon Got it. Thanks.
Fattorizzando per 2^a dentro la parentesi resta 1+2^a+(2^a)^2 e non (1 + 2^2 + 2^3)
❤
I gave up math in high school, but I realized how good the writing instruments you use are.😊
Wow...tough, but well explained
почему отказались от комплексного корня? Если при решении мы не отказались сразу от комплексных числе, то должны были воспользоваться понятием вектора. А вектор существует на плоскости, стало быть и комплексное решение этого уравнение СУЩЕСТВУЕТ!
Где на практике применяются комплексные числа? Никогда не понимал, зачем учиться делать дурную работу.
@@user-sr5lw3bv9 Дружище, если в твоей практике не применяются, то это не значит, что у всех. У инженеров в расчетах электрики, в гидравлике, аэродинамике ты шагу не ступишь без них.
Другое дело, что в ролике школьная задача, а в школе мнимые числа не используют
Задача для школьников, а не студентов ВУЗа.
Насколько я помню, в школе векторную алгебру не изучают и комплексные числа не используют.
@@user-sr5lw3bv9На самом деле комплексные числа применяются во многих областях, например, в электротехнике для описания колебательных процессов.
Просто автору ролика надо было либо указать, что при D
@@user-sr5lw3bv9For example in solving AC circuits? Complex numbers have many practical applications.
I have not found where you define if "a" is an integer, a real, a complex or else. Please for next problem precise it as soon as possible.
This is nit picking. But that is what math is all about. So, I definitely agree.
As stated, there are three solutions to the problem. And the answer given is not correct. It is an approximation, to three decimal positions (or four characters, if you wish). I believe that the correct answer should have stopped at 'a equal log of three to the base 2'.
Thank you and yours for your efforts. May you and yours stay well and prosper.
计算步骤写得太详细了，很多心算就能直接解决了。
So. Easy
How elegant the formulas dance while this wonderful music plays 🙂.
Olympiad Math exponential problem: 2ª + 4ª + 8ª = 39; a =?
a > 0, 8ª > 4ª > 2ª > 0; a ϵ R⁺, 2ª + 4ª + 8ª = 2ª + (2ª)² + (2ª)³ = 39
Let: b = 2ª; 2ª + (2ª)² + (2ª)³ = b + b² + b³ = 39, b + b² + b³ - 39 = 0, b = 2ª > 0
(b³ - 3³) + (b² + b - 12) = (b - 3)(b² + 3b + 9) + (b - 3)(b + 4) = (b - 3)(b² + 4b + 13) = 0
b² + 4b + 13 > 0, b ϵ R⁺; b - 3 = 0, b = 3 = 2ª, a = log₂3 = 1.585
The calculation was achieved on a smartphone with a standard calculator app
Answer check:
a = log₂3 = 1.585, 2ª = 3: 2ª + 4ª + 8ª = 2ª(1 + 2ª + 2²ª) = 3(1 + 3 + 9) = 39; Confirmed
Final answer:
a = log₂3 = 1.585
Not bad task. Not too complicated, but not so easy . Maybe little bit for logical combinations than real mathematic knowledge. Answer is 3/2
Completing the square would have been much easier than using the quadratic formula.
The music almost get me done 😅
令2^a=b
則原式：b+b^2+b^3=39
當b=2 則2+4+8=14
當b=3 則3+9+27=39
則2^a=3 則a=log3/log2
39=3+9+27,39=13*3,13=1+4+8
After t = 2^a, we have t+t²+t³=39. By guessing (factor of 39 is +-1, +-3, +-13), we get t=3
Then we divide t³+t²+t-39 by (t-3) to get t²+4t+13 and since D=4²-4*13
If only real solutions for t are wanted then really it's overkill to find the complex ones. From the discriminant, or even finding the turning point, it is clear this quadratic has no real roots.
Really very nice . 👍
Very educating combination instrument in the background ❤❤
And I liked the music.
Thank you. May you and yours stay well and prosper.
2^a=x , 4^a = 2^a * 2^a = x^2 , 8^a = 2^a * 2^a * 2^a = x^3 , 39 = 3 + 3^2 + 3^3
2^a + 4^a + 8^a = x + x^2 + x^3 = 3 + 3^2 + 3^3 => x=3 = 2^a
2^a=3 => a=ln3 / ln2 = 1,58496...
各国、綺麗とされる字がちがうのもいいよね！
1
exacty, about 1.5
Those complex t values were valid. They would have led you to a = ln(-2+-3i)/ln(2).
Or log2 3 (log base 2 of 3)
What's the logic behind writing 39 = 3 + 9 + 27?
Exponents of 3. 3=3 9=3^2 27=3^3 they did it to simplify
Сколько чернил и бумаги потрачено зря. Я решил это в голове за 15 секунд
Way too complicated. Notice right away 3 is a root and then use polynomial division to find t^2+4t+13 as the other factor.
2a + 4a + 8a = 39
14a = 39
a = log 39 / log 14
a = 1.38821
14^1.38821 = 39
Так решать нельзя. Там степень, а не умножение
log3/log2
I just fast forward the vid all the way to the end for the answer.
We don't want to use any formula,
See this easy method,
2^a+4^a+8^a=39
2^a+(2^2)^a+(2^3)^a=39
2^a+2^2a+2^3a=39
Here 2^a is taken a common,
2^a(1+2^2+2^3)=39
Adding this,
2^a(1+4+8)=39
2^a(13)=39
2^a=13
Taking log on both side
log 2^a=log 3
a log 2= log 3
So, a= log3/log 2
a= log base(2) 3 is the answer
I believe you have a mistake when you factored the 2^a out . When you factor 2^a out of 2^a + 2^(2a) + 2^(3a) the result is not 2^a(1 + 2^2 + 2^3) or 2^a(1 + 4 + 8). The result is 2^a(1 + 2^a + 2^(2a)). If your statement were true, then let's multiply your statement back and see if we can recover the original statement . . . 2^a(1 + 4 + 8) = 2^a(1 + 2^2 + 2^3) = 2^a + 2^a(2^2) + 2^a(2^3) = 2^a + 2^(a+2) + 2^(a+3)), and clearly that is not the same as 2^a + 2^(2a) + 2^(3a). Remember when you multiply like bases you ADD exponents!
Direto no excel dá 1,424 . 2^1,585 + 4^1,585 + 8^1,585 = 54 .
1.588 something
high school freshman level
13*2^a = 39
2^a = 3
end
slime rancher aah music
2ⁿ-1
그냥 보자마자 log2(3)
Is there a tradition to solve 2**a=3 as log 3/log 2 instead of simply log₂3 (by the definition of logarithm)?
You can, but the two most conventionally used logarithms are log of e and log of 10, especially log of e.
When in doubt, put it in log of e.
That cant be an Olympian problem. I could do that in my head.
You're then allowed to feel better about yourself
2^a=t is better
問題に「aが実数のとき」という条件がないのと、aの解はあくまでlogの近似値なのでおかしいのではないか。
Very noisy
a=log_2(3)でいいのでは？
SO EIN BlÖDSINN ! Wer braucht denn so was?
Easy. Olympian ? Are you sure?
Muy fácil, factorizando el 2^a salía más rápido
Considerando que (a^n)•(a^m)=a^(n+m), entonces si factorizas 2^a dentro del paréntesis te queda (1+(2^a)+(2^a)^2)), si pensaste que después de factorizar te quedaba (1+2^2+2^3) pues tuviste un error de conceptos ahí y llegaste solo por casualidad al resultado
Я там немного не понял - демона всё таки аызвали в конце ?
2^a + 4^a + 8^a = 39
so 2^a + 2^2a + 2^3a = 39
Let x = 2^a then
x"3 + x^2 + x - 39 = 0
but 27 + 9 + 3 = 3 and
x^3 - 3x^2 + 4x^2 - 12x + 13x - 39 =0
so x =3 or
x^2 + 4x + 13 = 0 in which case
x = (-4+-sqrt(16-52))/2 = -2+-6i
Математике надо запретить пользоваться буквами. Оставить только числа. Тогда математика будет математикой в не философией
最後が雑