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!
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.
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!
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.
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
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
Complex solutions shouldn't be rejected, unless it's specified, they're mathematically valid
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!
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!
From t³ + t² + t - 39 = 0, the rational root theorem will yield t = 3. Easy from there.
So. Easy
Beautiful music!
No reason given for rejecting complex solutions
The music almost get me done 😅
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)
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
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.
How elegant the formulas dance while this wonderful music plays 🙂.
log3/log2
Wow...tough, but well explained
What's the logic behind writing 39 = 3 + 9 + 27?
That cant be an Olympian problem. I could do that in my head.
You're then allowed to feel better about yourself
1
exacty, about 1.5
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!
Very educating combination instrument in the background ❤❤
SO EIN BlÖDSINN ! Wer braucht denn so was?
2ⁿ-1
Would it just be easier just to write log(base 2) of 3?
Easy. Olympian ? Are you sure?
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.
And I liked the music.
Thank you. May you and yours stay well and prosper.
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.
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.
Very noisy
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
Really very nice . 👍
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