The number of ordered pairs $(a,b)$ that are solutions for the equation $\log_{2^a}\left(\log_{2^b}\left(2^{1000}\right)\right)=1$ How many $(a,b)$ for $a,b \in \Bbb{N}$ pairs can satisfy the following equation:
$$\log_{2^a}\left(\log_{2^b}\left(2^{1000}\right)\right)=1$$
The answer is $3$, but I can't figure out how to get that answer.

This is my attempt.
$$\log_{2^a}\left(\log_{2^b}\left(2^{1000}\right)\right)=1$$
$$\frac{1}{a}\log_2\left(\log_{2^b}\left(2^{1000}\right)\right)=1$$
$$\log_2\left(\log_{2^b}\left(2^{1000}\right)\right)=a$$
$$\log_{2^b}\left(2^{1000}\right)=2^a$$
$$\frac{1}{b}\log_{2}\left(2^{1000}\right)=2^a$$
$$\log_{2}\left(2^{1000}\right)=2^ab$$
$$2^{1000}=2^{2^ab}$$
$$1000=2^ab$$
That's it! This is dead end.
Honestly, this is the best I could do altough I very much doubt that I can get two variables by solving one equation (for that we need a system of equations!). So, I think that I need another approach that will either give me what $a$ and $b$ can be or direct answer (i.e. the number of possible values for $a$ and $b$), but I don't know which one.

I agree with your derivation
$$\log_{2^a}\left(\log_{2^b}\left(2^{1000}\right)\right)=1\iff \log_{2^b}\left(2^{1000}\right)=2^a\iff (2^b)^{2^a}=2^{1000}\iff b\cdot 2^a=1000$$
now we can have


*

*$a=1, 2^a=2, b=500$

*$a=2, 2^a=4, b=250$

*$a=3, 2^a=8, b=125$
