วีดีโอ

Somewhat simpler to notice from 5^2a =50 that 5^(2a-2) =2. Then 2a-2=log(2)/log(5) and you're nearly done.

It's too easy 😅 a=1,b=2. It ' s not for genius

3.45

5^2a=50 2a*ln(5)=ln(50) a=ln(50)/ln(5)/2

Where does it say x+y needs to be natural only?

n^32=2^n =>n^1/n=2^1/32 =2^2^(1/64) =4^2^(1/128)=16^2^( 5:05 1/256) =256^1/256 =>n=256 =

Nice ❤❤❤❤❤

12/12

12/12

12/12

Lurs

Hi ☺ 7^sin²x + 7^cos²x=8 7^sin²x + 7^(1-sin²x)=8 {cos²x+sin²x=1} 7^sin²x + 7/7^sin²x=8 Let 7^sin²x= t t + 7/t =8 *t t²-8t=-7 t²-8t+16=-7 + 16 (t-4) ²= 9 t-4= +-√9 t=4 +-3 t=7 or t=1 Recall 7^sin²x= t Case1 : t=1 7^sin²x= 1 → 7^sin²x= 7⁰ sin²x=0 sinx sinx =0 sinx=0 Case2: t= 7 7^sin²x= 7 → sin²x=1 sin²x-1=0 (sinx+1)(sinx-1)=0 sinx=+-1 Answ: sinx=0 or sinx=1 or sinx=-1 → x=k Ꙥ/2 k is Z

You are writing Pi in a very strange way. "Only genius" is of course pure bullshit, I solved it right away from the thumbnail, half of it in my head

You are writing y like 4 and vice versa sometimes confusing the viewer.

X = 6

I think it's easier this way x^x^x^3=3 both sides ^3^3 x^x^x^3^3^3=3^3^3 { a^b^c=a^c^b } x^3^x^3^x^3=3^3^3 from here we get x^3=3 both sides ^(1/3) { a^b^c=a^(b*c) } x=3^(1/3)

You're getting wrong values for the inverse sine at 12:00, and later in the video you're not copying the same answers you got before. The correct answers for x are: ±π/3 + 2kπ, ±2π/3 + 2kπ, ±π/6 + 2kπ and ±5π/6 + 2kπ. The procedure is right though. It looks to me as if the person writing in the video is just copying from the real work but has no maths skills, for example when they write 304 instead of 30^4.... Nice video but you should've checked it before uploading it.

The problem is about addition, not multiplication, the very first step is wrong 2^a + 2^a + 2^a is not 2^3a 😂 😂

the original problem was a summation, 2^a + 2^a + 2^a. You solved for 2^a x 2^a x 2^a.

*_Could it be 0??_*

Math olympiad, which one? Clickbait. Thumbs down. That's an ordinary school excercise.

2^a+2^a+2^a is not 2^(a+a+a) WTF! I think you meant 2^a*2^a*2^a Get your math right 😆

For maths olympiad this is very simple !

7ˣ+7²ˣ+7³ˣ= 14 (7ˣ)+(7ˣ)²+(7ˣ)³= 14 7ˣ= y --> y³+y²+y= 14 y³+y²+y-14= 0 a,b,c ∈ ℤ --> y= t-b/3a (t-⅓)³+(t-⅓)²+(t-⅓)-14= 0 --> t³+⅔t-385/27= 0 t³+pt+q= 0, p,q ∈ ℝ By Cardano's Formula --> t= ⁷⁄₃ y= t-⅓ --> y= ⁷⁄₃-⅓= 2= 7ˣ xlog₇︎(7)= log₇︎(2) (y³+y²+y-14)/(y-2)= 0, y ∉ ℝ Solution: {𝔁 ∈ ℝ / 𝔁= log₇︎(2)} *EXTREME* *EASY!*

Once you have b(b²-3)=0, there are only two possibilities: b=0 and b²-3=0, where b=√3 and b=-√3 are almost directly deductable. In my opinion is an easier way. Good luck!

a-b-c-d = (a-b)-(c+d), not (c-d). That’s why x1 is correct and both x2 and x3 are lame. And it has nothing to do with Einstein, it’s kinda 5th grade at school.

I don't know how I did it orally 🤓

when right hand side value came to left hand side it should've been division instead of subtraction

I found 1.02239294

With such a slow music, very boring

You really don't need spell out in such laborious detail just how √(-4) = 2i. Anyone who's aware of the quadratic formula is well aware of how it produces complex solutions; and anyone who's not aware of the quadratic formula would have abandoned the video after the first minute. So next time, miss out the unnecessary extra steps.

In turkey this is a question for secondary school 😮

There is a mistake in the sign, that does not influence the result, but is rather ....

a⁴-(a-1)⁴ = 0 (a²)²-((a-1)²)² = 0 (a²+a²-2a+1)(a²-a²-2a+1) = 0 Let's set both factors equal to zero: 2a²-2a+1 = 0 a²-a+½ = 0 a²-a+¼ = -¼ (a+½)² = -¼ a=-½±½i a²-a²-2a+1 = 0 -2a+1= 0 a= ½

There is a very short way to the 3 solutions: x1= 1/2 is obvious , and if we write the equation as (a-1)^4= (± i)^4 * a^4 , the other two solutions x2= (1+i)/2 and x3 = (1-i)/2 follow immediately.

0

To get a solution (in the set N of numbers), there was a shorter path, it seems to me... x^32 = 2^x (x^32)^(1/32x) = (2^x)^(1/32x) x^(1/x) = 2^(1/32) recall: • (a^b)^c = a^(bc) and reciprocally a^(bc) = (a^b)^c • => 2^(1/32) = (2^8)^(1/32/8) x^(1/x) = (2^8)^(1/32/8) x^(1/x) = 256^(1/256) ■ x = 256 🙂

21/2

You don't solve the problem given in the title !

Me encantó el cambio de 25 por 35.... genial!!!

Clickbaits. You are far from being a Genius !😅😅😅

Man, why did you change it to 35?

The most unexpected knowledge I got from this video is that 25=35 😂 🤦‍♂️

Are you sure for x=0 , giving 0 in the denominator ?

ab+bc=ca b(a+c)=ca b=ca:(b+c) a(ac:(a+c))=100 a^2×c=100(a+c) a^2×c=100a+100c 300a=100a+100c 200a=100c 2a=c 2a^2=ca 2a^2=300 a^2=150 [a=\|150] b\|150=100 [b=100:\|150] c(100:\|150)=200 c=200\|150:100 [c=2\|150] a+b+c=\|150+100:\|150+2\|150=(150+100+300):\|150=550:\|150=44.9

This video is way way over complicated ab=100, ac= 300 => c=3b. : bc=200, ab=100 => c=2a => a=1.5b. Thus a+b+c = (1.5 +1 + 3) b = 5.5b : bc = 3 x b sq. =200 => b sq. = 200/3 = (2 x 100 ) /3 . Thus b = sq rt 2 / sq rt 3 x10 & a+b+c = sqrt2 /sq rt3 x10 x5.5 No pen or paper required.

Can't do it. I'm not that good at math.

I chanced that: a=100 // b=0 // c=200 ab=100+0 // bc=0+200 // ca=200+100 a+b+c=100+0+200 = 300 10 seconds of mental calculation instead of four A4 pages with calculations :)

Allo🎉 n est une puissance de 2 n=2^lp implique 2^n=2^32p implique p=2^(p_5) qui a pour solution p=8 qui me donne n=2^8=256🎉

I reasoned that a, b, and c were all the same sign, because ab, bc, and ca were all positive. So, when I got a^2, b^2, and c^2, I went ahead and assumed a = sqrt(a^2), b = sqrt(b^2), and c = sqrt(c^2), and figured a + b + c that way, knowing that the answer had to be plus or minus that. Usually, puzzles that look like this end up being something were you can't figure out a, b, and c individually, so you have to find some other trick to get to the sum. So this one was easier than I expected.