Відео

Let's see if there are integer solutions. Powers of 4 end in 4 or 6, powers of 5 end always in 5, so their sum can't be a power of 10. We'll have to go for LN also this time. Then I'll watch the video

@haiderlughmani Transforming the equation to { (20^x* 5^x * 4^x)/ 4^x} and reducing to ( 20^x* 5^x) = 100^x = 5 But x-result is different from your solution, it is log 5 (base100) = 0,349485 Why this simplification has a different result? Other observation: If 20^x^2 is calculated as (20^x)^2, the same log solution 0,349485 works. The above shown simplification can be a proof that the convention of calculating exponential from the right side is questionable.

One of the real solutions is the golden ratio.

Before watching: Looking at this visually, I can tell the real answer is 2 right off the bat; (-2)^4 = 2^4. There are also two complex solutions. To find those, we need to do a little work. Using the binomial theorem we get x^4-16x^3+96x^2-256x+256 = x^4. Subtracting x^4 from both sides yields -16x^3+96x^2- 256x+256 =0. We should do synthetic division, but first let's divide both sides by -16 so that the first coefficient is 1. That gives us x^3-6x^2+16x-16. Using synthetic division, we determine that X=2 is a factor. Thus, we can rewrite this as (x-2)(x^2-4x+8). We use the Quadratic Formula with the second factor. When we do so, the discriminant is negative. Thus, our other two roots are complex, not real. If looking only for real solutions, X=2 is the only solution. If accepting complex solutions though, the quadratic formula gives us (4±√-16)/2 = (4±4i)/2 = 2±2i Real solution: X=2 Complex solutions: x = 2±2i

괄호도 제대로 못 맞추는 수준이라니...

Elegant solution

Did I miss the condition that x and y were integers ? Otherwise there are an infinite number of hyperbolic solutions.

I just solved by trial and error brute force since the equation yielded an integer.

I didnt like they way could be other way more easily

Very good!

Сложмтьоба уравнения получить х+у равно +7 и-7.

we get , x^3 - 6x^2 + 16x - 16=0 , (x-2)(x^2-2x+8)=0 , x=2 , +1 -2 x^2-2x+8=0 , x= 1+i*V7 , 1-i*V7 , -4 +8 test , x=2 , (2-4)^4=16 , 2^4=16 , OK , +8 -16=0 ,

Simpler way is a = 7^cos^2x and sin^2x = 1 - cos^2x This gives a^2-8a+7=0

2^(3a)=27 , let u=2^a , u^3 +/- u^2 +/- u - 27=0 , (u-3)(u^2+3u+9)=0 , u=3 , 2^a=3 , a=log3/log2 , +1 -3 u^2+3u+9=0 , u=(-3+/-V(9-36))/2 , u=(-3+/-i*V27)/2 , +3 -9 test , 8^(log3/log2)+8^(log3/log2)+8^(log3/log2)=27+27+27 , --> 81 , OK , +9 -27=0 ,

Hopefully you would use ,(comma) carefully. Comma means and, so x-1=0,x-7=0 is not proper expression. X-1=0 or x-7=0 would be correct. Mathematics is always related with logics, so 'and' and 'or' must be used correctly.

We really only need to find two powers of nine such that when one of them is subtracted from the other, then the answer is 6560. We also know that as the powers of nine increase, so does the difference between these powers. We can just try a few small powers under six and we find 9⁴ - 9⁰ = 6560 and we are done as x = 13 and y = 9. Why six? As 9⁶ - 9⁵ = 472,392 and the differences just keeps increases if we change six to seven.

Didática zero.... Nem dá para acompanhar....

Before watching: Alright, so right off the bat we can discard the chances of any *real* solution with (x-9)<4. Why? Because 9^4 is the first power of 9 that gives you a number remotely near 6560; specifically, it gives you 6561. 6561-9^(y-9) = 6560 -> 1 = 9^(y-9), meaning that y-9 = 0 (as 9^u =1 only for u=0), so Y = 9 Thus, (x,y) = (13,9).

Cách này không hay. Dài dòng. Không cần tính cụ thể x, y. Cộng 2 phương trình ta được phương trình 3. Nhân 2 phương trình ta được phương trình 4. Lấy phương trình 4 chia phương trình 3 tìm được xy = 28.21/28+ 21= 12

Multiply both sides by 64, let u=4m. u^3-4u+3=0 has a clear solution of 1, or m=1/4. This leaves u^2-3u-3=0. Then u=(3+/-sqrt(21))/2, for m=(3+/-sqrt(21))/8.

😢ты маньяк твою мать ❗

Ez nem megoldás. Ez nem más, mint a megsejtés igazolása. Szornyű, hogy ez a tudálékos okoskodás kering.

m= 1/4

m^2--(m^3/2)^2=4/64--1/64=(1/4)^2--(1/8)^2

Х 4. У 3

Very good!

9^(x - 9) - 9^(y - 9) = 6560 ← as the result is a positive number, (x - 9) > (y - 9) → x > y [9^(x) * 9^(- 9)] - [9^(y) * 9^(- 9)] = 6560 [9^(x) * {1/9^(9)}] - [9^(y) * {1/9^(9)}] = 6560 {1/9^(9)} * [9^(x) - 9^(y)] = 6560 9^(x) - 9^(y) = 6560 * 9^(9) → recall: x > y 9^(x + y - y) - 9^(y) = 9^(9) * 6560 9^(y + x - y) - 9^(y) = 9^(9) * 6560 [9^(y) * 9^(x - y)] - 9^(y) = 9^(9) * 6560 9^(y) * [9^(x - y) - 1] = 9^(9) * 6560 → as 6560 is not divisible by, you can deduce that: 9^(y) = 9^(9) → y = 9 [9^(x - y) - 1] = 6560 9^(x - y) = 6561 9^(x - y) = 9^(4) x - y = 4 x = 4 + y → x = 13

Very clever rewiting the fractions that way.

Very good!

Все решается гораздо быстрей и проще

Mera naam kyon hai😮😮😮😮😮

m^2+m^3=5,625. m=?

(25)^1/2^log2•(25)^1/2 =5^log10 5^log10=5^1<=> α=(25)^1/2 or α=5

36=27+9=3^3+3^2 ?

Ek or method se banega dono eqn ko divide kar dete or phir se dono eqn ko add ka dete

Given X^2 +x y = 28--------1 Y^2 + XY =21---------2 Summing 1 & 2 x^2+2xy+y^2=49 (X + Y) ^2 = 49 X + Y = +7 or -7 Substitute this in 1 X(x+y) =28 X x 7 = 28 or X = 4 X x -7= 28 or X =-4 Substitute x value in eqn 2 Y(X + Y) = 21 Y( X + Y) =21 Y (+7 or -7) =21 Y = 21/ 7 =3 or 21/-7 = -3 So XY = 4 x 3 = 12 or XY = -4 x -3 = 12

Best solution ❤😊

الحل غير مثالي لأنك وضعت س=2

ليس حلا مثاليا لأنك فرضت أن س=2

Well...I got the first solution by inspection m = 1/4 Then I used (m-a)(m-b)(m-1/4) = 0 to get a cubic. Turn the resulting equation negative to get a coefficient of -1 for the m³ term. Next, set the coefficient of the m² term to be 1 and set the constant term equal to -3/64. You then solve two equations for the unknown "a" and "b", getting the final two solutions of m = (3±sqrt(21))/8

If you use the LambertW function you get two solutions x=0,5111965.. and x=128 4^x=(2x)^32 (2x)ln2=32ln(2x) ...etc... -ln(2x)=W(-(ln2)/32) x1=(1/2)*e^[-W0(-(ln2)/32)]=0,5111965.. x2=(1/2)*e^[-W1(-(ln2)/32)]=128

I thought the unknowns were sine x and cosin x. I solved this in a different way and got the same answer and faster, but I still like your method. I think you should have mentioned what is the unknown and also once you find that sin^(x) =+/-1, 0, you should use the first trig identity to find the value of the cosine^2(x). The whole idea was to apply algebra in a quick way to arrive to he conclusion. Keep up the good work.

64m²-64m³=3 ou encore 4(4m)²-(4m)³=3. On pose M=4m alors 4M²-M³=3. M³-4M²+3=0. 1 est une solution évidente. La division par M-1 est plus simple.

u^2+u-20=(u-5)(u+4)

You are making a 13 minutes long video about this equation and ignoring complex solutions. I would find the complex solutions.

(u+1)^4+(u-1)^4=2 u^4 + 12 u^2 + 2=706 2 u^4 + 12 u^2 - 704 = 0 u^4 + 6 u^2 - 352 = 0 v=u^2 v^2 + 6 v - 352 = (v+3)^2-9-35=(v+3)^2-361=(v+3)^2-19^2=(v+3+19)(v+3-19)=(v+22)(v+16)=(u^2+22)(u^2-16)=(u -i sqrt(22))(u +i sqrt(22)(u-4)(u+4)=0

y=x+9 kexit

b1=-8

全然、分かりません！ 大学で数字を専攻されたのですか？

m² - m³ = m²(1 - m) = 3/64. My first guess is that m is of the form 1/2ⁿ. More specifically, if 1 - m = 3/4 , then m would be 1/4 and this works! m = 1/4 is a solution, so dividing (m - 1/4) into m² - m³ - 3/64 = 0 and we find the quadratic factors and we can solve that with the quadratic formula.