Why did nobody mention x = -5, and y = -6? Then: x³ - y³ = (-5)³ - (-6)³ = -125 - (-216) = -125 + 216 = 91

3^(x³) / 9^(x) = 81 3^(x³) / (3²)^(x) = 3^(4) 3^(x³) / 3^(2x) = 3^(4) 3^(x³) * (3)^(- 2x) = 3^(4) → recall: n^(a) * n^(b) = n^(a + b) 3^(x³ - 2x) = 3^(4) x³ - 2x = 4 x³ - 2x - 4 = 0 x³ - (4x - 2x) - 4 = 0 x³ - 4x + 2x - 4 = 0 (x³ - 4x) + (2x - 4) = 0 x.(x² - 4) + 2.(x - 2) = 0 x.(x + 2).(x - 2) + 2.(x - 2) = 0 (x - 2).[x.(x + 2) + 2] = 0 (x - 2).[x² + 2x + 2] = 0 (x - 2).[x² + 2x + 1 + 1] = 0 (x - 2).[(x² + 2x + 1) + 1] = 0 (x - 2).[(x + 1)² + 1] = 0 → [(x + 1)² + 1] ≠ 0 → because: (square + positive number) > 0 (x - 2) = 0 x - 2 = 0 x = 2

Watching from Philippines sir

3^(m + 2^(m) = 35 3^(m) + 2^(m) = 27 + 9 3^(m) - 27 + 2^(m) - 9 = 0 → let: 3^(m) = a³ a³ - 27 + 2^(m) - 9 = 0 → suppose that: 2^(m) = a² a³ - 27 + a² - 9 = 0 a³ - 3³ + a² - 3² = 0 (a³ - 3³) + (a² - 3²) = 0 → recall: (x³ - y³) = (x - y).(x² + xy + y²) (a - 3).(a² + 3a + 9) + (a² - 3²) = 0 → recall: (x² - y²) = (x + y).(x - y) (a - 3).(a² + 3a + 9) + (a + 3).(a - 3) = 0 (a - 3).[(a² + 3a + 9) + (a + 3)] = 0 (a - 3).[a² + 3a + 9 + a + 3] = 0 (a - 3).[a² + 4a + 12] = 0 (a - 3).[a² + 4a + 4 + 8] = 0 (a - 3).[(a² + 4a + 4) + 8] = 0 (a - 3).[(a + 2)² + 8] = 0 → [(a + 2)² + 8] ≠ 0 → because a square + positive number > 0 (a - 3) = 0 a = 3 → recall: 3^(m) = a³ 3^(m) = 27 3^(m) = 3^(3) m = 3

Easier. 2^10 =1,024. 1,024*1,024=1,048,576. 1,048,576*2=2,097,152. Subtract 5 = 2,097,147. Recused sum of digits is 3 which is divisible by 3. 2,097,147/3=699,049. Reduced sum of digits =1, so only divisible by 3 one time. Test of prime divisors quickly reveals 13 (one time only), 699,049/13=53,773. Square root of 53,773 by long division ~ <232. So a test of primes less than 232 and greater 16 would show any others. .... there are none. So, 2^21 - 5 = 2,097,147 = 3 * 13 * 53,773.

First trial: 1/14 + 1/b = 1/13, ∵ 1/b = (14 - 13)/(13 × 14) = 1/182, ∴ a + b = 14 + 182 = 196 👈 For a ≠ b. Otherwise, 1/26 + 1/26 = 1/13 ⇒ a + b = 52 ✖

Eccezzionale!!!!!!

A 3 B 1

Law √a, fast ln ^ has to equal part of non partisan graph p y=0 10 parts only 11 is non category as equation deliberates ln function uv strand as √ has a I unit.. answer 12

Without constraints, the first solution is obvious x = y = 16. A second less obvious lower bound solution when y = 0 is x =(2*log(39)/log(8))^2. With a family of other non-integer solutions, where y>0, x = (2*lg(39+(sqrt(5))^Sqrt(y))/lg(8))^2

-ln3/9=-ln3*x*e^(-ln3*x) , /* 3/3 left side , -3*ln3/3^3=-ln3*x*e^(-ln3*x) , -3*ln3*e^(-3*ln3)=-ln3*x*e^(-ln3*x) , -3*ln3==-ln3*x , x1=3 , test , OK , W(-ln3/9)=~ -0.140478921 , x2= -0.140478921/-ln3 , x2=~ 0.127869 , test , OK ,

Thailand, Math Olympiad: (x + 6)⁴ + (x + 4)⁴ = 82; x =? First method: 82 > (x + 6)⁴ > (x + 4)⁴ > 0; 0 > x > - 6 or 82 > (x + 4)⁴ > (x + 6)⁴ > 0; - 6 > x > - 8 (x + 6)⁴ + (x + 4)⁴ = 81 + 1 = 3⁴ + 1⁴ = (- 3 + 6)⁴ + (- 3 + 4)⁴; x = - 3 (x + 6)⁴ + (x + 4)⁴ = 1 + 81 = 1⁴ + 3⁴ = (- 7 + 6)⁴ + (- 7 + 4)⁴; x = - 7 Missing two complex value roots Second method: Solve the equation directly to find the two complex value roots Let: y = x + 5; (x + 6)⁴ + (x + 4)⁴ = (y + 1)⁴ + (y - 1)⁴ = 82 (y ± 1)⁴ = y⁴ ± 4y³ + 6y² ± 4y + 1, (y + 1)⁴ + (y - 1)⁴ = 2(y⁴ + 6y² + 1) = 82 y⁴ + 6y² + 1 = 41, y⁴ + 6y² - 40 = (y² - 4)(y² + 10) = 0, y² - 4 = 0 or y² + 10 = 0 y² = 4, y = ± 2 = x + 5, x = - 3 or x = - 7; y² = - 10, y = ± i√10, x = - 5 ± i√10 Answer check: x = - 3 or x = - 7: (x + 6)⁴ + (x + 4)⁴ = 82; Confirmed as shown in First method x = - 5 ± i√10, x + 5 = ± i√10 = y: (x + 6)⁴ + (x + 4)⁴ = 2(y⁴ + 6y² + 1) 2(y⁴ + 6y² + 1) = 2[(± i√10)⁴ + 6(± i√10)² + 1] = 2(100 - 60 + 1) = 82; Confirmed Final answer: x = - 3; x = - 7; Two complex value roots, if acceptable; x = - 5 + i√10 or x = - 5 - 5i√10

3^x=9x=3^2.x 3^x/x=3^2 3^(x-1)=3^2 x-1=2 x=3 vérifier 3^3=9.3=27

無聊

👍

a,bが自然数だとは書いていない。

Ans :: a+b=52(a=26, b=26) a+b=196,(a=14,b=182, vice versa also)

Unneccessary fumbling about

1:10 Actually (a - b)^2 = a^2 + b^2 - 2ab. 6:47 Also, isn't sqrt(a^2) = +/- sqrt(-64) ?

Que questão bonita. Parabéns. Brasil Novembro de 2024.

1/4^x=x. , x4^x=1 taking ln lnx4^x=0. , lnx=-2xln2 lnx/x=-2ln2=2ln2^-1=ln2^-1/2^-1 result. , x=2^-1=1/2

I suppose we are looking for integer solutions right? Otherwise 1/a=1/13-1/b => a=13b/(b-13) if a,b>0 Then we get a<0 when 13>b>0 and the function drops from +inf asymptotic to 1. a+b=13b/(b-13)+b= (13b+b^2-13b)/(b-13)=b^2/(b-13)=b + 13/(b-13) which has infinitely many solutions over R If we are searching for integers removing the ratios makes sense so multiplying by 13ab we get: 13b+13a=ab ab-13b-13(a-13+13)=0 (a-13)(b-13)-169=0 (a-13)(b-13)=(+-1)*(+-169) Or (+-13)(+-13) => (a,b) And (b,a) € {(14,182),(12,-156),(26,26)} Ps check this nice video about Egyptian fractions from numberphile kzbin.info/www/bejne/l4e4hpWEl7aemLMsi=KSvOBCOzVoIBTujQ

Case 1. a=0, but a>0. Nothing

Condition: x > 0 Easy to see that 1 is a result. Why dont use the logarit function, it will be better and easier way to find 9/4 as the other result.

‏‪2^√a=x , log2^√a=logx. , √a=logx/log2 x+logx/log2=20. , logx/log2=20-x x=2^(20-x) , x=16 , 2^√a=x=16=2^4 , √a=4 , a=16

Good

✓x=b , x=b^2 , (b^2)^(b^3)=(b^3)^(b^2) haw 2 case case one b^3 =b^2. ,b>0 ,b=1 , x=1 case 2 are opposite now taking ln 2b^3lnb=3b^2lnb. , 3b^2=2b^3. , b=3/2. x=9/4

No need to calculation they are 1 and 9 (9+(9×1)+1)=19

Taking ln on both sides, the solutions (1 and 9/4) are more easily obtained. I think.

It is one liner. Original equation easily transforms to X^(X^(1/2) -3/2)=1 that has 2 solutions: 1) X = 1 and 2) X=(3/2)^2 = 9/4 (X^2-3/2 =0 ==> X=(3/2^2) = 9/4)

Faltan 2 posibles soluciones mas x=1 y x=0 y hay que probarlas...??

U R the Man, I love the way u explain it

Почему больше а больше 0? Должно быть? Вы упускаете корни уравнения а=0, б=19, проверьте и всё правильно будет 0+19×0+19=19!!!

x = 1

Good lesson

✓y=b , y=b^2 , ✓x=a , x=a^2 then (a^2)^b-(b^2)^2=17 , (a^b-b^a)(a^b+b^a)=17 =1×17 , a^b-b^a=1 and a^b+b^a=17 , a^b=9=3^2 and b^a=8=2^3 , a=3 , b=2 and x=3^2=9 , y=2^2=4

At the point where a²-(1/2a)=0 We have difference of two squares. Or? Couldn we have used the identity (a-b)(a+b)?

We can notice that x<=2 so inside radical exists √(2-x)<=2 so x>=-2 so outside radical exists X>=0 Also x=1 is a solution to equation 1=√2-√(2-1) Now by squaring x^2=2-√(2-x) √(2-x)=2-x^2 squarig again 2-x=4-4x^2+x^4 X^4-4x^2+x+2=0 (we know x-1 is solution) R x^4. X^3. X^2. X. 1 1. 0. -4. 1 2 1. 1. 1. -3. -2 0 2. 1. 3. 3. 4. Not a root -2. 1. -1. -1 0. root So we have x^2-x-1=0 or x=1 or x=-2 rejected (<0) X3=(1+√5)/2 x4=(1-√5)/2 rejected as it is <0 If we are using x3 then √(2-x)=√(3-√5)/2=√(6-2√5)/4=1/2(√5-1) Then the outer root is √2-(√5-1)/2=√(5-√5)/2 does not satisfy equation

Amazing. I messed up calculations trying some substitutions but intorducing a square was impressive. Still after substituting a and b i woukd not square just use the power of 1/2 as square root getting √x^√y=9=3^2 √y^√x=8=2^3 Wonder if this is sufficient to get the solutions or will itnkead to omission

(2a+1)b=44-a^2 . 2a+1>=3 b=(44-a^2)/(2a+1)=(1/4)*{(-2a+1)+175/(2a+1)} (2a+1)(2a+4b-1)=175 (2a+1,2a+4b-1)=(5, 35),(7, 25)

(2a+1)b=44-a^2>0.a=1,2,3,4,5,6. (a,b)=(2,8),(3,5)

You a pro and an excellent teacher . THANK YOU!!!

a shall > 0

Muito bom. Parabens

5/a + 6/b =7 => 1+6=7 => 5/a=1 , a=5 => 6/b=6 , b=1 => 5+2=7 => 5/a=5 . a=1 => 6/b=2 , b=3

X=4

Essa eu achei difícil

mi solucion solo mirando la miniatura m=-3 ya las otras posibles soluciones las dejo para el video mi primera vista mi opcion fue el 3, pero como luego vi que el m^2 estaba antes del m^3 supuse que era un numero negativo, total m^2 siempre dara como resultado un numero positivo y al m^3 quedar negativo daria como resultado -(-x) lo cual da otro numero positivo y solo queda sumarlos XD

In3^x-2=lnx then (x-2)ln3=1lnx then x-2=1 and 3=x then x=3

Why do you calculate again? If a+1=5, then a=4, isn't it?