Thanks...
I've thought of another way, but am not sure if it's valid. Please let me know if it's correct: -
$\displaystyle log_2(log_2(log_2(x))) = 1$
$\displaystyle log_2(log_2(log_2(x))) = log_2(2)$ [I can write $\displaystyle log_2(2)$ instead of $\displaystyle 1$, can't I?]
Yes.
So,
$\displaystyle log_2(log_2(x)) = 2$ [Now this is what I'm unsure about. Can I eliminate $\displaystyle log_2$ from both sides in this case??]
Yes you can. This is because the function $\displaystyle \color{red}f(x)=\log_2(x)$ is one-to-one, or injective, which you probably haven't learned yet. But it means that f(a) = f(b) implies a = b, and graphically, f(x) passes the "horizontal line test."
Next,
$\displaystyle log_2(log_2(x)) = 2*log_2(2)$ [where $\displaystyle log_2(2) = 1$ ]
$\displaystyle log_2(log_2(x)) = log_2(2)^2 \Longrightarrow log_2(log_2(x)) = log_2(4)$
I would write $\displaystyle \color{red}log_2(2^2)$ so people don't misinterpret as $\displaystyle \color{red}\left(\log_2(2)\right)^2$.
$\displaystyle log_2(x) = 4$
$\displaystyle \Longrightarrow log_2(x) = 4*log_2(2)$
$\displaystyle \Longrightarrow log_2(x) = log_2(2)^4$
$\displaystyle \Longrightarrow log_2(x) = log_2(16)$
Thus $\displaystyle x = 16$
Now is it a valid method???
Yes. Good job. 