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=2^11+(2^11)^2+(2^11)^3 /(2^11)^3-1 Let 2^11=y y^3+y^2+y/y^3-1 =y(y^2+y+1)/(y-1)(y^2+y+1) =y/y-1 =2^11/2^11 -1 X= 2048/2048-1=2-48/2047/.

I would approach it with analysis creating a function and finding the max and min in every interval instead of using w functions since it is made for high school students

just for reference default tactics here is to guess solutions, and then prove (via derivatives) that all solutions found, and most llm doing it that way, but actually you can force gemeni 2 thinking into doing via lambert function, and it does it more carefully than you: " This is the analytical solution expressed in terms of the Lambert W function. To find the specific values of x, we need to evaluate W(-ln(2)/2). Since -ln(2)/2 ≈ -0.346, which is between -1/e and 0, there are two real values for W(-ln(2)/2), corresponding to its two real branches: W₀ and W₋₁."

x=1 is also solution, so you did lose 50% of solutions.

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1/2 x + 1+ tan^2x -12=1/cot^2x cosec^2x =1/sin^2x// ,sec^2x =1/cos^2x 1/2x

10^m(3)/5^m(2)=96 10^m(3)(2)/5^m(2)(3)=96(2/3) 10^m/5^m =96(2/3) (10/5)^m= 32x2 2^m =2^5.2^1 2^m=2^5+1 m=6//

x=-3//

log3(2).log(3)/log(2) = log10 (2)/log10(3) .log3/log10(10) = log10 (2) =log2/log3.log3/1 =log2 =log2/log2=1/.

=1..

1/(2y)^3 =64 1/y^3=64x8=512 3√y^3=√1/512 y^3=1/8//

(2^3x+3^3x)/(2^2x.3^x+3^2x.2^x)=7/6

Very interesting..

How long did it take people in the Olympics to get this one? For me it took 30 second ish to solve

3^m-2^n=3^5-2^5 n=5 , m=5

📊♥️

We can also form this equation. x² + xy = 22 ---(i) y² + xy = 99 ---(ii) Add (i) and (ii) x² + xy + xy + y² = 22+99 x² + 2xy + y² = 121 (x + y)² = 121 x + y = √121 x + y = 11 Then we substitute the value of x as you did. Then we can find y. After that we substitute value of y to find x.

Спасибо! Очень полезно для мозга! Вспомнила школьную программу!👍

or 2

its -2

Let's raise both sides to the -⅓ power 2y = 1/³√64 2y = 1/4 y= 1/8

When it is explained like that, the problem becomes straightforward. Nice video!

wtf are u doing, just replace that m with 6

Does log 7 to the base i exist?

3

x=42 ta gueule

ERRORS!

I'm into negative and positive mathematics

(9-x)^x=2^7 so x=7

2

Se tivesse substituído por a e b no início ficaria bem mais simples. Chegaria em a = 3b. (If I had replaced it with a and b at the beginning it would be much simpler. It would arrive at a = 3b.)

10^0.5~3.16; LHS is 6.16^x + 0.16^x; Calculator: try x=2>37.97. close enough. Check video. Darn! I missed the -2 solution.

F(x) - x^3 = - 3x^2 x^2 - 3x^2 = - 2x^2 - 2x^2 = 80 F'(x) - 2x^2 = - 4x - 4x = 80 x = 80/- 4 x= - 20 Demostrate. - 3*2*(-20) - 40 = 80

Girice 200g + Hambie - 3.50 euro, Papaline 200g + Hambie 120g (or 80-100g, I don't know, but I mean 120g is standard - 1.20 Euro, and Fulie 240g is standard - 2.00 euro) - 3.00 Euro, Hambie - 0.50 Euro. Can to buy only 1 Hambie per 200g. Is that good?!

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Nice

Thank you sir 🙏⭐

Note that x = sqrt(3) - 1 is the solution to the equation x^2 = 2 - 2x, so x^8 = (x^2)^4 = (2 - 2x) ^4, expand and use x^2 = 2 - 2x as needed and you will eventually get x^8 = 16 (41 - 56x), substitute x = sqrt(3) - 1 and you will get the same result :)

Why not?: (√3-1)^8 =((√3-1)^2)^4=(3-1)^4=2^4=16?

Thanks a lot

Love this, looks perplexing to start with but is surprisingly simple.

Well done!

Really cool, can you show intergrade W function by the way

Grande spiegazione ma anche grande difficoltà nel leggere : poteva usare un pennarello che rendesse la scrittura visibile !

4095=4096-1=64^2-1^2=(64-1)(64+1)=(x^3-1)(x^3+1) x^3=64=4^3 x=4. 1/4^(-x)=4^x=4^4=256.

Excellent really enjoyed and understood your explanation. Thank you very much.

(x² + 9) (x + 3) (x − 3) = (x² + 9) (x² − 9) = x⁴ − 81 ³√(x⁴ − 81) = ³√(424 − 81) = ³√343 = 7

X^256 =4^x Find x

you made it so complicated

x = 12