Відео

Convince me with better things 'cause this doesn't count a thing. Try harder before you deserve to be shown anything. 😮

the answer is 2

Nicely done. I like the new way you have adopted.

Nice use of identity, very clear.

x=log18/log6=(log6+log3)/log6=1+log3/(log2+log3) x≈1+0.477/(0.301+0.477)≈1255/778≈1.61311 Deviation is about 0.00004 6^1.61311≈17.999

(2x-3y)⁵ = 1×(2x)⁵ - 5×(2x)⁴×(3y)¹ + 10×(2x)³×(3y)² - 10×(2x)²×(3y)³ + 5×(2x)¹×(3y)⁴ - 1×(3y)⁵ = 32x⁵ - 240x⁴y¹ + 720x³y² - 1080x²y³ + 810x¹y⁴ - 243y⁵

i love this trick so much its my fav math thing

The cross check was so satisfying

x = 2?

Nice 👍🏽

10&8 ?

Your method is incorrect because X^2 +2x -3x =0 X=-3 x=1 I use. Viyet theory

x²+2x+1-4=0 (x+1)²=4 x+1=±2 First root -2-1= -3 Second root 2-1= 1

-3²-(-3)³=9-(-27)=9+27=36 So x= -3

A great solution

-3

Yet again proving Olympiad math just means make it a million times more complicated then it needs to be.

Thanks for the challenge, which I solved in my head. However, I was only looking for real solutions - although the question did not specify that only real solutions are required.

Coincidental. Determine x for x^x^x=17

Keep it up

This problem could be solved without factoring. Just get your constant on one side and all variable terms on the other. And when you do that, 0 is not even a possible answer anyway. It can only be 1/144.

That's just the quadratic formula in disguise but with extra work

Нищо особено. Интересно... но за начинаещи. Фактически е модификация на метода с отделяне на точен квадрат.

Nice? This is a Joke.!

-3

Seriously? This is an Olympiad problem?

X^2-x^3 =36, x^2 -x^3= 3^2 + 3^3, x=3

X=9

Very clear.

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Ok, solved by a simple observation. But how would one go about formally deriving the solution?

Una manera nada elegante de demostrar la igualdad

why arent there 5 solutions

Is it really Stanford university's exam? Too easy. In Japan, it is junior high school exam.

thats a really long solution, you can just do t=sqrt of x and than a=1 b=sqrt 5 and c=-4, much faster espacially if you use a calculator

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Great video and tutorial but this is far too complicated for an Oxford University entrance exam

I solved it using iteration of x = (2ln7)/(ln x) it approximates to 3.278

This is just a direct application for Lambert W function . the answer is x = 3.278

Any complex solutions?

4ˣ = 8^2^^3 4ˣ = 8^[2^(2²)] 4ˣ = 8^(2⁴) 4ˣ = 8^(2¹·2⁴⁻¹) 4ˣ = (2³)^(2·2³) (aᵐ)ⁿ = aᵐ*ⁿ = (aⁿ)ᵐ (2²)ˣ = (2²)^(3·2³) x = 3·2³ x = 3·8 x = 24

Bro just use log6 on both sides and you get the answer in one step

(x^3/4)(x^1/8)/(x^3/4) = 3^3/4 Cancel and raise to the 8th power (x^1/8)^8 = x = (3^3/4)^8 = 3^6=729

👍🎉

How many hours u study ?

😂😂

Here’s how I would rationalize. If the equation is nearly identical it had to be 0. Idk I feel like logic solved this one faster

How x2 equals to zero sir