Spain l can you solve this exponential problem? l k=? l Olympiad Mathematics.

[@generalbushdakta1856]

The task should state the domain of K, if K is a real number or if it also could be a complex number.

[@leonidkuznetsov8240]

If K is integer then K={-2,2}, if k is real then K=2 and if K is complex number there are 6 roots

[@iulianmihai9367]

k=2

[@АлексейКоженов-т4ж]

Или - 2.

[@BartOlomeo-cr5gm]

k= -2

[@UlricGrim]

K^6 = 2^6 K^6 = 64 6root(K^6) = 6root(64) K = 2 and -2 When you take the root of both sides you get a (+) and (-) answer. I’m a simple man.

[@charlescomerlatto3936]

just raise both sides to the power of 1/6 and you get two answers: +2/-2.

[@normanstewart7130]

What about the other four solutions?

[@jacekskrzymowski6715]

​​@@normanstewart7130they must be on the vertices of a hexagon on the complex plane.

[@Soudchaser91q41]

K=2 ?

[@АлексейКоженов-т4ж]

Зачем так сложно? Если к в 6 степени равно 2 в 6 степени, то 2 или - 2..

[@dr.m.k.hardaha4114]

K^6=2^6 [K^6]/[2^6]=1 [K/2]^6=1 [K/2]=1^(1/6) [K/2]=1 K=2

[@БелАлекс]

Уравнение 6 степени всегда имеет 6 ответов. Вы написали один, где остальные 5?

[@AnilSonkar-ly5ex]

Nice video

[@payoo_2674]

k^6 = 2^6 k^6 = 2^6 * 1 1 = e^(2nπi) n∈Z (k^6)^(1/6) = (2^6 * e^(2nπi))^(1/6) k = 2 * e^(kπi/3) n = 0, 1, 2, ..., 5 for n=0: k₁ = 2*e^(0*πi/3) = 2*(cos(0)+i*sin(0)) = 2*(1+i*0) = 2 for n=1: k₂ = 2*e^(1*πi/3) = 2*(cos(π/3)+i*sin(π/3)) = 2*(1/2+i*√3/2) = 1+√3i for n=2: k₃ = 2*e^(2*πi/3) = 2*(cos(2π/3)+i*sin(2π/3)) = 2*(-1/2+i*√3/2) = -1+√3i for n=3: k₄ = 2*e^(3*πi/3) = 2*(cos(π)+i*sin(π)) = 2*(-1+i*0) = -2 for n=4: k₅ = 2*e^(4*πi/3) = 2*(cos(4π/3)+i*sin(4π/3)) = 2*(-1/2-i*√3/2) = -1-√3i for n=5: k₆ = 2*e^(5*πi/3) = 2*(cos(5π/3)+i*sin(5π/3)) = 2*(1/2-i*√3/2) = 1-√3i

[@alfredofettuccine9425]

Using Euler’s identity, K=2*e^niPi/3

[@DmitriNesterov]

How many different roots here? Really all six?

[@alfredofettuccine9425]

@@DmitriNesterov Yes, of course. n=0, K=2 n=1, K=2e^iPi/3=2cos(Pi/3)+2i*sin(Pi/3)=1+i*sqrt(3) n=2, K=-1+i*sqrt(3) n=3, K=-2 n=4, K=-1-i*sqrt(3) n=5, K=1-i*sqrt(3) n=6, you have gone full circle and are back to 2…

[@DmitriNesterov]

@@alfredofettuccine9425 большое спасибо! Чувствую, что должно быть очевидно, но картинка в голове не складывается. Старею, совсем обленился :( Спасибо! :-) ❤️ P.s. но я интуитивно чувствовал, что Ваш ответ - правильный, просто, уточнил 😊

[@prabirsarkar504]

Basically it is not a math at all