I fucked around, and this is what i got:

🍔 + 🍔 + 🍔 = 18

=> 🍔 = 6

🍔 + ( 🍟 × 🍟 ) = 5

🍟² = -1

=> 🍟 = ±i

(🥤^ 🍟) -🥤= 3

note:

e^(x+iy) = (e^x)(cos(y) + i sin(y))

case 1: (🥤^ i) -🥤= 3

case 2: (🥤^ -i) -🥤= 3

at this point, i just used wolfram to get both:

=> {

🥤 ≈ -2.97983 + 0.0388569 i,

🥤 ≈ 0.27972 - 0.748461 i

}

Curiously, case 1 and case 2 return the same 2 values…