# A problem: help needed

1. Nov 16, 2006

### murshid_islam

this is the problem:
let {ak} 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?

2. Nov 16, 2006

### Office_Shredder

Staff Emeritus
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