Can You Pass Massachusetts Institute Entrance Examination?

[@tulsaken2754]

How do you find out what the lambert function of a constant is? I don’t have W on my calculator.

[@tommyliu7020]

Wolfram Alpha.

[@MoonRocks77]

Between 2 & 3, about 2.47

[@paolovolante]

AI Claude 3.5 Sonnet would pass the exam.
We start with the equation: 2^x + x = 8
Rearrange to: 2^x = 8 - x
Multiply both sides by 2^(-x):
1 = (8 - x) * 2^(-x)
Multiply both sides by -1/8:
-1/8 = (x/8 - 1) * 2^(-x)
Let y = x/8 - 1, so x = 8(y + 1)
Substituting:
-1/8 = y * 2^(-8(y+1))
Multiply both sides by -8:
1 = -8y * 2^(-8(y+1))
1 = -8y * (1/2^8)^(y+1)
1 = -8y * (1/256)^(y+1)
This is in the form of z = we^w, where:
z = -1/(256ln(2))
w = -8yln(2)
The solution to z = we^w is w = W(z), where W is the Lambert W function
So: -8yln(2) = W(-1/(256ln(2)))
Solve for y:
y = -W(-1/(256ln(2)))/(8ln(2))
Remember that y = x/8 - 1, so:
x/8 - 1 = -W(-1/(256ln(2)))/(8ln(2))
Solve for x:
x = 8 - W(-1/(256ln(2)))/ln(2)
x = 8 - W(1/(256ln(2)))/ln(2) (using the property W(-x) = -W(x))
Simplify:
x = 8 - W(256*ln(2))/ln(2)

[@too_ql]

😮

[@darkslayer5444]

impressionnant j'apprécie fortement ton travail !

[@justinwooten9998]

a^x + x = b, x = b - W(a^b × ln(a))/ln(a), a ≠ 1, a & b > 0, x > 0.

[@RealQinnMalloryu4]

{2x+2x ➖ }+{x +x ➖ }={4x^2+x^2}=4x^4 2^2x^2^2 1^1x^1^2 x^1^2 (x ➖ 2x+1).

[@richardcommins4926]

My 10 second guess was somewhere between 2 and 3 so I guessed around 2.5 done.

[@robotnikkkk001]

...AND MAYBE IT'D BE BETTER TO MAKE DIFFERENTIAL
........BECAUSE THIS W IS SUCH A GIMMICK

[@PrakritiPandey-bi3jt]

It is more easy to solve by using graph

[@simonlevett4776]

No i could not pass the entrance exam, but who gives a damn ?

[@eureoma578]

Bond.... James Bond ln2

[@1990ner1]

Jeez that W constant 😄. Thanks for the lesson

[@canalMatUem]

Legal a funcão W de lambert

[@momiezz9934]

the answer is 2.468

[@simonlevett4776]

Thank you Einstein.

[@KLONDYKE1111]

you speak too fast

[@haierpad5669]

What a nonsense, return me my 11 minutes please...