Відео

Совершенно дикое решение. И ещё и логарифм без основания!

a= .5

Подбориком на изи раскатал

Draw circle of radius 5 centered on the origin of the complex plane. Mark points on circle at 120° intervals starting with x=5.

1st ❤❤

eˣˡⁿˣ = eˡⁿˣ (x - 1)lnx = 0 *x = 1*

Much simpler in polar coordinates. r, theta

80=64+16 . From Greece. This way is more easy.

From Greece

Why did you take 3. 3 does not devide 80.

One thing I like to tell you , whenever you are teaching math put the division line where equal sign is there. I am teaching math since 1962. I have seen that most of the teacher are unaware of it.

a=2:b in a²+b² --> b² + b Wurzel 2 - 2 = 0 --> b = 2,3 --> a = 0,86 --> 2,3 x 0,86 = 1,999 ~ 2

1

Bro saying this a quick and viral method 😂😂😅

by Vieta theorem， it's able to directly conclude that u and v is two roots of the equation z^2 - 8z + 8 = 0.

I have no idea why people ignore Argand diagrams and Euler's formula: e^(iθ) = cosθ + i sinθ x¹²=1=e^(i2πn) for n=0,1,2,3,... 11 (repeats thereafter) x=e^(iπn/6) giving four roots on the (Argand diagram) axes (1,-1,i,-i) and two sets of four roots based on 30,60,90 triangles : ±√3/2±i/2 and ±1/2±i√3/2

u and v are roots of x² - 7x + 7 = 0 (b = sum = 7 and c = 7 = product) => x1 = u = (7 + sqrt(7² - 4·7))/2 ≈ 5.791 and x2 = v = (7 - sqrt(7² - 4·7))/2 ≈ 1.208

Cube root 2 also apparent post factoring parent left side x^3(x^3-1)

tbh x=-1 is pretty apparent from the parent problem lol

A Nice Math Olympiad: 8ˣ + 2ˣ = 130; x = ? 8ˣ + 2ˣ = (2ˣ)³ + 2ˣ = 130 = 125 + 5 = 5³ + 5; 2ˣ = 5, x = log₂5 = 2.322 The calculation was achieved on a smartphone with a standard calculator app Answer check: 8ˣ + 2ˣ = 130; Confirmed as shown Final answer: x = log₂5 = 2.322

ab/(a+b) = 1/(1/a+1/b) = 1/(log7/log441 + log3/log441) = 2

You say, 97% failed to solve?! So, Germany has so low level of education?! Poor Germany...

You can simply solve quadratic equation x*x-7x+7 by standard Vieta formula.

You made it way harder than it needs to be. What a terrible solution... Just do this: x^64 = 4^x log(x^64) = log(4^x) 64 log(x) = x log(4) 64 log(x) = 2x [use base 2 logs] 32 log(x) = x 32u = 2^u [let u = log(x)] 2^5 u = 2^u u = 2^(u-5) Now obviously u > 5, so let's try u=6.... u=7.... u=8 !Bingo! So 8 = log(x) x = 2^8 = 256

There is a mistake in the problem 3rd step (1-1=0)😊

i'm 11 years old

Valde Valde facilis! Hic a= 7 est. Responsi.

I used your method to the transformation x^3+x^2=7^3+7^2. That leads to x=7. The rest ist polynomial division and solution of the quadratic equation.

take the log both side from the beginning , we have 2xlog2=(2--x)log3 , and then .....

Math Olympiad: 9ˣ⁺¹ - 9ˣ⁻¹ = 20; x = ? 9ˣ⁺¹ - 9ˣ⁻¹ = (9ˣ)(9 + 1/9) = (9ˣ)(80/9) = 20; 9ˣ⁻¹ = 1/4, (3ˣ⁻¹)² = (2⁻¹)² √[(3ˣ⁻¹)²] = √[(2⁻¹)²], 3ˣ⁻¹ = 2⁻¹; Let: 3ª = 2, a = log₃2, 3ˣ⁻¹ = 2⁻¹ = 3⁻ª x - 1 = - a, x = 1 - a = 1 - log₃2 = 1 - 0.631 = 0.369 Answer check: x = 0.369: 9ˣ⁺¹ - 9ˣ⁻¹ = 9¹·³⁶⁹ - 9⁻⁰·⁶³¹ = 20.25 - 0.25 = 20; Confirmed The calculation was achieved on a smartphone with a standard calculator app Final answer: x = 1 - log₃2 = 0.369

A 5, B 8,C 7 is the answer

x = 1 or x = -ln(15)/ln(2) And that can be calculated on any calculator (x = -1.68260619449).😎

X= 1.

same funds , 5^(ln3/ln5)x * 5^x^2 = 5^(ln15/5) , (ln3/ln5)x + x^2 = (ln15/5) , x= 1 , -1.68261 , test x=1 , 3^1 * 5^1 =3*5 , 3*5=15 , same , OK , test x= -1.68261 , 3^(-1.68261)*5^((-1.68261)^2)=14.9999... ~ 15 , OK ,

I think there is no point doing all of that and not solving for complex values. At the start you assume that one of the solutions will be 7. a^3 + a^2 = 392 a^3 + a^2 = 343 + 49 a^3 + a^2 = 7^3 + 7^2 that is it

Yes, or ... 3^x·5^(x^2) = 15 notes: • log₅(3) = 0,682606 • log₅(15) = 0,682606 5^(0.682606x)·5^(x^2) = 5^1,682606 5^(0.682606x + x^2) = 5^1,682606 same base (= 5) => 0.682606x + x^2 = 1,682606 x^2 + 0.682606x - 1,682606 = 0 Δ = 0.682606² - 4·1·(-1.682606) = 7.196374 √7.196374 = ±2.682605 • root #1: x = (-0.682606 + 2.682605)/(2·1) = 0.99999 = 1 • root #2: x = (-0.682606 - 2.682605)/(2·1) = -1.682605 ----- final results ----- ■ x = 1 ■ x = -1.682605 🙂

Its faster to guess and adjust repeatedly

x = 1/3

999×999

2

5 7 8

7 8,5

My solution would be ▶ 4¹⁺ˣ + 4¹⁻ˣ = 10 4*4ˣ + 4*4⁻ˣ= 10 4*4ˣ + 4/4⁻ˣ= 10 both sides multiplied by 4ˣ : 4*4²ˣ + 4= 10*4ˣ let 4ˣ = u ⇒ 4u²+4= 10u 4u²-10u+4=0 2u²-5u+2=0 Δ= 25-4*2*2 Δ= 9 √Δ= 3 ⇒ u₁= (5+3)/2*2 u₁= 2 4ˣ = 2 2²ˣ = 2¹ 2x= 1 x₁= 1/2 u₂= (5-3)/2*2 u₂= 1/2 4ˣ = 1/2 2²ˣ = 2⁻¹ 2x= -1 x₂= -1/2 𝕃 = { x ∈ ℝ | (-1/2, 1/2) }

Good video. This was real helpful.

u and v are interchangeable, obviously. This suggests two solutions for each with u₁ = v₂ and u₂ = v₁. So we expect a quadratic function: u + v = 8 uv = 8 u + v = uv = 8 u = 8 − v (8 − v) v = 8 −v² + 8v − 8 = 0 v² − 8v + 8 = 0 v₁,₂ = 4 ± √(16 − 8) v₁ = 4 − 2√2 ∨ v₂ = 4 + 2√2 u₁ = v₂ = 4 + 2√2 ∨ u₂ = v₁ = 4 − 2√2 Check: u₁ + v₁ = 4 + 2√2 + 4 − 2√2 = 8 u₁v₁ = (4 + 2√2) (4 − 2√2) = 4² − (2√2)² = 16 − 8 = 8 𝕃 = {(4 − 2√2, 4 + 2√2), (4 + 2√2, 4 − 2√2)}

Or x = 1 and y = 0

Let a=1 a+2ab+b=1 1+2b+b=1 b(3)=0 b=0 a=1 b=0 or b=1 a=0

10⁵⁰

Don't you think it should be non negative integers and not positive integers, as positjve integers doesn't include 0. Otherwise, great video sensei.

Questão muito inteligente! Obrigado Professor.