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เข้าร่วมเมื่อ 28 ก.ค. 2021
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Japanese l can you solve this?? l Olympiad Mathematics
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n<4. n>2. If n is a positive integer, n=3.
)4+4)2
You solved clearly thanks
No wonder
9^×-x^4=65 3^×)^2-×^2)^2=65 3^×+×^2) (3^×-×^2)=13(5) =>3^×1+×^2=13 & 3^×-×^2=5 3^×=9=3^2....&×=2 ×^2=4=2^2.....×=2
A=1
X = 2 is the ANSWER this is a Problem to be solved Mentally
Moroccan !can you salve thist??il faut demander au marocains aussi , ils sont forts en maths
a to the power 0 isn’t 1, it’s a
Its soo easy 😭🙏🙏 brodda why make it hard 😭😭😭💀
When you get paid by the hour
How on earth are you supposed to answer this quickly and efficiently?
3
Pra que facilitar se pode complicar muito né? De cabeça A=1
More simply: a is one of dividers of 3 (1,3), and b is one of dividers of 4 (1,2,4), is only 2 solutions for equality 3/a+4/b=5 : (a=1, b=2) and (a=3 , b=1)
Log of a negative number ? Impossible.
Where is the negative number here?
ChatGPT says only -512i.................check
Wow
Very good!
n=3
3*5^a=3*4*5. 5^(a-1)=4=2²=2^(lb5+a-1). lb5+a-1=2. a=2+1-lb5=3-lb5.
Seems to be unworthy for use on Maths Olympiads. Too simple.
What is i ?
a=1
Very good!
❤
Цікаво 😊
HAY UN ERROR -1024 AL pasar al otro lado cambia de signo a + debió sumar sin embargo UD testa
Great work with beautiful background music.
Wrong
x=log(5) 20 x=1.8614
i do not agree. The suggested "substitutions" in all 4 cases are like AxB=CxD and from here is derived a new equality --- A=C and B=D, which is wrong! Let's substitute few numbers --- 2x10=4x5 - neither 2 is equal to 4, nor 10 to 5! And anything after that does not make any mathematical sense.
The method is too complicated!
There is mistake at 03:15.
At 4:15 instead of -4X i write here -2X.
A mistake on 4:17. -8x -2x = -10x, not -12x
Here will be -4X. Mistakenly written -2X.
Add a comment on the video
mesmerizing.
m+m^2+m^3 =13 next m(1+m+m^2)=3*13 for m = 3 we get 3^a = 3^1 that mean a = 1 for ( 1+ m +m^2 = 13 ) m^2 + m - 12 = 0 (m-3)(m+4)=0 m-3 = 0 gives that m = 3 and a = 1 m+4 = 0 gives m = -4 here we don't have real solution .
The sum of three terms equal to 39raisetoa aeqalto one
Compañero fuiste en total fracaso. Demasiados comentarios de dándote soluciones mucho más rápidas.
What's happening here
Very good!
Good @MATHS TUTORIAL (BL SAHU)
Method is unnecessarily complicated. 4^2a=48 Take ln both sides: 2a*ln4=ln(4*4*3) 2a*ln4=2ln4+ln3 Divide both sides by 2ln4 a=1+ln3)/2ln4 But ln4=2ln2 a=1+ln3/4ln2
Entirely agree.
Test the answers, prove it
very good
Exasperatingly slow
I prefer the notation 81 \times (-1). Brackets avoid any misunderstanding.
Including complex roots: 4^a * 4^a = 48 4^(a + a) = (48^[1 / 2])^2 4^(2 * a) = ([2^4 * 3]^[1 / 2])^2 4^(a * 2) = ([2^4]^[1 / 2] * 3^[1 / 2])^2 (4^a)^2 = (2^[4 * (1 / 2)] * 3^[1 / 2])^2 (4^a)^2 = (2^2 * 3^[1 / 2])^2 (4^a)^2 = (4 * 3^[1 / 2])^2 sqrt([4^a]^2) = +/- sqrt([4 * 3^(1 / 2)]^2) 4^a = +/- 4 * 3^(1 / 2) (1 / 4) * 4^a = +/- (1 / 4) * 4 * 3^(1 / 2) 4^a * 4^(-1) = +/- 4^1 * 4^(-1) * 3^(1 / 2) 4^(a - 1) = +/- 4^(1 - 1) * 3^(1 / 2) 4^(a - 1) = +/- 4^0 * 3^(1 / 2) 4^(a - 1) = +/- 1 * 3^(1 / 2) 4^(a - 1) = +/- 3^(1 / 2) (2^2)^(a - 1) = +/- ([3^(1 / 2)]^[1 / 2])^2 2^(2 * [a - 1]) = +/- (3^[(1 / 2) * (1 / 2)])^2 2^([a - 1] * 2) = +/- (3^[1 / 4])^2 (2^[a - 1])^2 = +/- (3^[1 / 4])^2 (2^[a - 1])^2 = +(3^[1 / 4])^2, or (2^[a - 1])^2 = -(3^[1 / 4])^2 Let x = 2^(a - 1), and y = 3^(1 / 4) (2^[a - 1])^2 = (3^[1 / 4])^2, or (2^[a - 1])^2 = -(3^[1 / 4])^2 => x^2 = y^2, or x^2 = -y^2 => x^2 - y^2 = y^2 - y^2, or x^2 + y^2 = -y^2 + y^2 => x^2 - y^2 = 0, or x^2 + y^2 = 0 => (x - y)(x + y) = 0, or (x - i * y)(x + i * y) = 0 => (2^[a - 1] - 3^[1 / 4])(2^[a - 1] + 3^[1 / 4]) = 0, or (2^[a - 1] - i * 3^[1 / 4])(2^[a - 1] + i * 3^[1 / 4]) = 0 Suppose 2^(a - 1) - 3^(1 / 4) = 0 2^(a - 1) - 3^(1 / 4) = 0 2^(a - 1) - 3^(1 / 4) + 3^(1 / 4) = 0 + 3^(1 / 4) 2^(a - 1) = 3^(1 / 4) ln(2^[a - 1]) = ln(3^[1 / 4]) (a - 1) * ln(2) = (1 / 4) * ln(3) (a - 1) * ln(2) / ln(2) = ln(3) / (4 * ln[2]) (a - 1) * log_2(2) = log_2(3) / 4 (a - 1) * 1 = log_2(3) / 4 a - 1 + 1 = log_2(3) / 4 + 1 a = log_2(3) / 4 + 1 Suppose 2^(a - 1) + 3^(1 / 4) = 0 2^(a - 1) + 3^(1 / 4) - 3^(1 / 4) = 0 - 3^(1 / 4) 2^(a - 1) = -3^(1 / 4) ln(2^[a - 1]) = ln(-1 * 3^[1 / 4]) ln(2^[a - 1]) = ln(i^2 * 3^[1 / 4]) (a - 1) * ln(2) = ln(i^2 * 3^[1 / 4]) (a - 1) * ln(2) / ln(2) = ln(i^2 * 3^[1 / 4]) / ln(2) (a - 1) * log_2(2) = ln(i^2) / ln(2) + ln(3^[1 / 4]) / ln(2) (a - 1) * 1 = ln(e^[i * tau / 2]) / ln(2) + log_2(3^[1 / 4]) a - 1 = (i * tau / 2) * ln(e) / ln(2) + log_2(3^[1 / 4]) a - 1 = (i * tau / 2) * 1 / ln(2) + log_2(3^[1 / 4]) a - 1 = i * tau / (2 * ln[2]) + (1 / 4) * log_2(3) a - 1 + 1 = i * tau / (2 * ln[2]) + log_2(3) / 4 + 1 a = i * tau / (2 * ln[2]) + log_2(3) / 4 + 1 a = i * 2 * tau / (2^2 * ln[2]) + log_2(3) / 4 + 1 a = i * 2 * tau / (4 * ln[2]) + log_2(3) / 4 + 1 Suppose 2^(a - 1) - i * 3^(1 / 4) = 0 2^(a - 1) - i * 3^(1 / 4) = 0 2^(a - 1) - i * 3^(1 / 4) + i * 3^(1 / 4) = 0 + i * 3^(1 / 4) 2^(a - 1) = i * 3^(1 / 4) ln(2^[a - 1]) = ln(i * 3^[1 / 4]) (a - 1) * ln(2) = ln(i * 3^[1 / 4]) (a - 1) * ln(2) / ln(2) = ln(i * 3^[1 / 4]) / ln(2) (a - 1) * log_2(2) = ln(i) / ln(2) + ln(3^[1 / 4]) / ln(2) (a - 1) * 1 = ln(e^[i * tau / 4]) / ln(2) + (1 / 4) * ln(3^[1 / 4]) / ln(2) a - 1 = (i * tau / 4) * ln(e) / ln(2) + ln(3) / (4 * ln[2]) a - 1 = (i * tau / 4) * 1 / ln(2) + ln(3) / (4 * ln[2]) a - 1 = i * tau / (4 * ln[2]) + ln(3) / (4 * ln[2]) a - 1 = i * tau / (4 * ln[2]) + log_2(3) / 4 a - 1 + 1 = i * tau / (4 * ln[2]) + log_2(3) / 4 + 1 a = i * 1 * tau / (4 * ln[2]) + log_2(3) / 4 + 1 Suppose 2^(a - 1) + i * 3^(1 / 4) = 0 2^(a - 1) + i * 3^(1 / 4) = 0 2^(a - 1) + i * 3^(1 / 4) - i * 3^(1 / 4) = 0 - i * 3^(1 / 4) 2^(a - 1) = -i * 3^(1 / 4) ln(2^[a - 1]) = ln(-1 * i * 3^[1 / 4]) ln(2^[a - 1]) = ln(i^2 * i * 3^[1 / 4]) ln(2^[a - 1]) = ln(i^3 * 3^[1 / 4]) (a - 1) * ln(2) = ln(i^3 * 3^[1 / 4]) (a - 1) * ln(2) / ln(2) = ln(i^3 * 3^[1 / 4]) / ln(2) (a - 1) * log_2(2) = ln(i^3) / ln(2) + ln(3^[1 / 4]) / ln(2) (a - 1) * 1 = ln(e^[i * 3 * tau / 4]) / ln(2) + (1 / 4) * ln(3) / ln(2) a - 1 = (i * 3 * tau / 4) * ln(e) / ln(2) + ln(3) / (4 * ln[2]) a - 1 = (i * 3 * tau / 4) * 1 / ln(2) + ln(3) / (4 * ln[2]) a - 1 = i * 3 * tau / (4 * ln[2]) + log_2(3) / 4 a - 1 + 1 = i * 3 * tau / (4 * ln[2]) + log_2(3) / 4 + 1 a = i * 3 * tau / (4 * ln[2]) + log_2(3) / 4 + 1 a1 = i * 0 * tau / (4 * ln[2]) + log_2(3) / 4 + 1 a2 = i * 1 * tau / (4 * ln[2]) + log_2(3) / 4 + 1 a3 = i * 2 * tau / (4 * ln[2]) + log_2(3) / 4 + 1 a4 = i * 3 * tau / (4 * ln[2]) + log_2(3) / 4 + 1
Great!!! Thanks!!!!
No. I stopped at 1'10" because you made a mistake. (√a) ² not equal a, but | a |.. Also, you should know that to solve an equation, one must know the definition set.. 🙄