([e ^ v] of ((2) * ([ln v] of (3)))) // 3^2 = 9

([e ^ v] of ((0.5) * ([ln v] of (9)))) // 9^0.5 = sqrt(9) = 3

([e ^ v] of (([ln v] of (9)) / (2))) // 9^(1/2) = sqrt(9) = 3

awesome-llama wrote:

You can combine the e^ and ln math operators to achieve exponentiation (or to find the nth root).

Just some examples, feel free to pick or modify whatever suits your needs…

([e ^ v] of ((2) * ([ln v] of (3)))) // 3^2 = 9

([e ^ v] of ((0.5) * ([ln v] of (9)))) // 9^0.5 = sqrt(9) = 3

([e ^ v] of (([ln v] of (9)) / (2))) // 9^(1/2) = sqrt(9) = 3

([10^ v] of (([log v] of (...)) / (...))::operators)

qwerty_wasd_gone wrote:

It doesn't have to be that, it can also be base 10 logarithms.

([10^ v] of (([log v] of (...)) / (...))::operators)

case 'ln': return Math.log(n);
case 'log': return Math.log(n) / Math.LN10;
case 'e ^': return Math.exp(n);
case '10 ^': return Math.pow(10, n);