How to express any integer using 3 twos?

Question

Four arithmetic operations, logarithms and square radicals √
are allowed.

For example 0 = √(2 + 2) - 2.

Answer 1

Any positive integer n may be represented thus:

-ln (ln (√√√√...√2)/ln 2)/ln 2

Where the square root symbol is iterated n times.

Why this works: √2 = 2^(1/2), √√2 = (2^(1/2))^(1/2) = 2^(1/4), √√√2 = 2^(1/8), and in general √√√√√...√2 (n square roots) = 2^(1/2^n). Now, ln x/ln a = log_a x, so ln (√√√√...√2)/ln 2 = log_2 (2^(1/2^n)) = 1/2^n. And therefore -ln (ln (√√√√...√2)/ln 2)/ln 2 = -ln (1/2^n)/ln 2 = -log_2 (1/2^n) = -log_2 (2^(-n)) = -(-n) = n. Omitting the negative sign allows you to express negative integers in the same way. You've already given an expression for 0, or alternatively, you can simply use this expression with no square roots (i.e. -ln (ln (2)/ln 2)/ln 2 = 0). Thus any integer whatsoever may be expressed using three twos.

Answer 2

nowise: 100000000000 cannot be expressed for sure