Finding the value of n, so that it is bigger than M? We introduce some notation for writing really big (but finite) numbers.
A googol, denoted g, is defined by $g = 10^{100}$. A googolplex, denoted G, is defined by
$G = 10^g$. A MathPatharoo, denoted M, is defined by $M = G^G$.
Next, for positive integers m, n define m ↑ n (spoken “em uparrow en”) as follows:
$$m ↑ 1 = m$$
$$m ↑ (n+1) = m^{m↑n}$$
And, find the smallest positive integer value of n for 2↑n, so that $2↑n \ge M$
After trying some values for n in 2↑n, the value is always 2 raised to itself $n-1$ times. For example, 2↑2 has a value of $2^2$, 2↑3 has a value of $2^{2^2}$, 2↑4 has a value of $2^{2^{2^2}}$, and so on. I do not know how to approach finding a value of n for 2↑n, so that $2↑n \ge M$ because the numbers are all way too big. Thanks for the help!
NOTE: The ↑ is not any other mathematical term and doesnt relate to any other mathematical expression. It is just an expression made up for this specific problem. Also, I must show exactly on paper how this answer was derived. 
 A: We have
$$M = (10^{10^{100}})^{10^{10^{100}}} = 10^{10^{100}10^{10^{100}}}
= 10^{10^{100+10^{100}}} \lt 10^{10^{10^{101}}} \\
\lt 2^{4\cdot 10^{10^{101}}}
\lt 2^{64^{10^{101}}}
\lt 2^{2^{6\cdot 10^{101}}}
\lt 2^{2^{10^{102}}}
\lt 2^{2^{2^{4\cdot 102}}}
\lt 2^{2^{2^{2^{512}}}}
\le 2^{2^{2^{2^{9}}}}
\le 2^{2^{2^{2^{2^4}}}}
\le 2^{2^{2^{2^{2^{2^2}}}}}
= 2\uparrow\uparrow 7
$$

We can also approach it the other way:
$2\uparrow\uparrow 2 = 4$
$2\uparrow\uparrow 3 = 16$
$2\uparrow\uparrow 4 = 65536$
$2\uparrow\uparrow 5$ is astronomically larger than a googol. In particular it is much larger than $\log_2(10)$ googols. But it is less than a googolplex.
$2\uparrow\uparrow 6$ is therefore much larger than a googolplex -- in fact, easily larger than a googol googolplexes.
Since your $M$ is approximately $10^{\text{googolplex}}$, $2\uparrow\uparrow 7$ is therefore larger than $M$.
A: $M = G^G = 2^{G\log_2 G} = 2^{G^2 \log_2 g} = 2^{100 G^2 \log_2 10}\approx  2^{332.19 G^2}$.
$2 ↑ k = 2^{...}\ge 2^{332.19 G^2}$
$k-1 \ge 332.19 G^2$
$k-2 = \log_2 332.19 G^2= 2*8.375 \log_2 G = g*2*8.375 \log_2 g = 100*g*2*8.375 \log_2 10 = 100*g*2*8.375 *3.3219$
$k-3 = 2*3.3219 * 100*3.3219*\log_2 8.375*\log_2*3.3219$
So $k = \lfloor 2*\log_2 10*100*\log_2 10*\log_2(\log_2 10*100)*\log_2 10 + 3 \rfloor$
