Finding the last term in sequence with a_n = log₂(a_{n-1})

[murshid_islam]

this is the problem:
let {a_k} be a sequence of real numbers. and the sequence is defined as

$$a_0 = 2006$$

$$a_{n} = \log_{2}a_{n-1}$$

now i have to find out the number of terms in the sequence.

this is what i have done:
i see that a₅ is negative. so we can't find out a₆ or any other subsequent terms. so the number of terms is 6 (a₀ upto a₅).

but how can i find out the number of terms without calculating the terms upto a₅? is there a way to do it?

 

[Office_Shredder]

If $$a_{n} = \log_{2}a_{n-1}$$ is true, then $$2^{a_n} = a_{n-1)$$

So this tells us $$2^{2^{2...^{2^{a_n}}}} = a_0$$ So find how many times you can put two to itself before it goes past 2006.

I think that works