Evaluating tetration to infinite heights (e.g., $2^{2^{2^{2^{.^{.^.}}}}}$)

The Problem

How can you evaluate (i.e., get a value for) Tetration (i.e., iterated exponentiation) to infinite heights?

For example, what would be the value of this expression?

$$ 2^{2^{2^{2^{2^{.^{.^.}}}}}} $$

My (pathetic) Attempts

I tried equating it to $x$ and substituting it on the RHS, but no luck:

$$ x = 2^{2^{2^{2^{2^{.^{.^.}}}}}} $$ $$ x = 2^x $$ $$ x = \log_{2}{x} $$

What I think we need to do is have the RHS as a polynomial with one variable and the RHS a constant so we can solve for $x$.

I tried drawing the equation on Wolfram Alpha but the lines on the graph don't touch, so no luck there either.

Novice mathematician here. Thanks.

Edit

Sorry, I am a dolt :(

I didn't realize this was a diverging series. What confused me is my math sir told me it could be done. What he actually meant was that it could be said like this:

$$ \frac{\log{x}}{x} = \log{2} $$

but I somehow assumed there would be a numerical answer.

Alternative Questions

@Clayton's answer suggested an similar question which was a convergent series. While that wasn't what my sir meant, it practically could've been:

$$ \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}} $$

Another one I can think of would be:

$$ \sqrt{2*\sqrt{2*\sqrt{2*\sqrt{2*\sqrt{...}}}}} $$

Anyway, interesting question this has turned out to be...

Solution 1:

Hint: Equations $y=x$ and $y=2^x$ do not intersect, means there's no solution for $x\in\mathbb R$.

Solution 2:

Here are the first 20 solutions (for $\log(z)+k \cdot (2\cdot \pi\cdot i)$ such that $2^z=z$ or $z\cdot \log(2) = \log(z) + k\cdot(2 \pi i)$ : $$ \small \begin{array} {r|l} k & z : 2^z=z\\ \hline 0 & 0.82467854614+1.56743212385 \, î \\ 1 & 3.51523672192+10.8800532084 \, î \\ 2 & 4.36143141283+20.0871628060 \, î \\ 3 & 4.88885664543+29.2211855083 \, î \\ 4 & 5.27384865880+38.3277872288 \, î \\ 5 & 5.57736047492+47.4208762811 \, î \\ 6 & 5.82797084936+56.5062285608 \, î \\ 7 & 6.04142622483+65.5867042057 \, î \\ 8 & 6.22733941446+74.6638922661 \, î \\ 9 & 6.39201436790+83.7387507973 \, î \\ 10 & 6.53981045480+92.8118939379 \, î \\ 11 & 6.67386852707+101.883734634 \, î \\ 12 & 6.79652672953+110.954561406 \, î \\ 13 & 6.90957275012+120.024582285 \, î \\ 14 & 7.01440413644+129.093951274 \, î \\ 15 & 7.11213417750+138.162784952 \, î \\ 16 & 7.20366413122+147.231173287 \, î \\ 17 & 7.28973387243+156.299186877 \, î \\ 18 & 7.37095826487+165.366881942 \, î \\ 19 & 7.44785382986+174.434303834 \, î \end{array} $$

(Using Pari/GP , more than 100 digits precision)

l2=log(2)
pi2i = 2*Pi*I 
{list=matrix(20,2);
for(k=0,20-1,              \\ k contains branchno for logarithm
   x0=1+I;
   for(j=1,20,              \\ Newton-iteration
        x1=x0-(l2*x0-(log(x0)+k*pi2i))/(l2-(1/x0));
        if(abs(x1-x0)<1e-100,break(),x0=x1); );
   list[1+k,]=[k,x0];
 );}
 printp(list)