Logarithms of logarithms of Graham's number, is the result ever handy?

Just getting $3 \uparrow\uparrow\uparrow 3$, which is a power tower of $3 \uparrow 3 \uparrow 3=3^{27}\ \ =7625597484987\ \ 3$'s to be a handy number takes $7625597484985$ applications of the $\log$ to get to $3^3=27$. The logarithm is woefully inadequate for this purpose.

The concept of $\log^*$ is a step in the right direction, but still not enough. We have $\log^* 3 \uparrow\uparrow\uparrow 3=7625597484985$, which isn't handy, but $\log \log^* 3 \uparrow\uparrow\uparrow 3=27$ is. Unfortunately we have a lot more uparrows to go. We probably need to define $\log^{**}$ as the number of times you apply $\log^*$ to get handy, then $\log^{***}$, etc. I suspect we need another (several) layers-define $\log^\&$ as the number of stars you have to put on $\log$ to get a handy number in one go. I have no idea how to do the computation, or even what sort of data structure is appropriate.

Can anyone estimate or even calculate how often to apply a log (of base 10, or 3 or e or any handy base) to arrive at a handy number?

Graham's number $G$ can be written as an exteremely tall exponential tower of $3$s:

$$G \ =\ 3\uparrow\uparrow height $$

With $3$ as the base of the logarithms, you're asking for the $height$ of this tower (or rather $height - 2$, for a handy number of $27$). Now, a formula for the exact height can be found using the following property of Knuth arrows:

$$3\uparrow^x y = 3\uparrow^{x-1}(3\uparrow^x(y-1)) \ \ \ \ (x \ge 2, y \ge 2)$$

Applying this repeatedly gives

$$3\uparrow^x 3 \\ =3\uparrow^{x-1}(3\uparrow^{x}2) \\ =3\uparrow^{x-2}(3\uparrow^{x-1}(3\uparrow^{x}2 - 1))\\ =3\uparrow^{x-3}(3\uparrow^{x-2}(3\uparrow^{x-1}(3\uparrow^{x}2 - 1) - 1)) \\ \cdot\\ \cdot\\ =3\uparrow^2( 3\uparrow^3 (3\uparrow^4 ( \ \cdots \ (3\uparrow^{x-1}(3\uparrow^{x}2 - 1) - 1) \cdots -1 ) ) ) $$

Thus, the handy number 27 is the result of starting with $G$ and applying the base-3 $\log$ exactly $height-2$ times, where $$height = 3\uparrow^3 (3\uparrow^4 ( \ \cdots \ (3\uparrow^{g_{63}-1}(3\uparrow^{g_{63}}2 - 1) - 1) \cdots -1 )) $$

NB: A very crude lower bound on $height$ is given by
$$3\uparrow^{g_{63}} 3 \ = \ 3\uparrow^{g_{63}-1}(3\uparrow^{g_{63}-1}3) \ \ggg \ 3\uparrow^{2}(3\uparrow^{g_{63}-1}3)$$

namely,

$$height \ \ggg \ 3\uparrow^{g_{63}-1}3 $$

which has only one less arrow than $G$ itself!