Expert: Robi Bhattacharjee - 9/10/2008

Question
Hi, how do you find the nth term in this logarithm sequence:
 log[base 2]8, log[base4]8, log[base8]8, log[base16]8, log[base32]8? Can you please show the steps? Thank you very much!!!

Answer
Hello Emily,

Well basically, you notice that each term in this sequence is pretty much the same except for the base (it is a 'log' and it is of 8). So lets look at the bases of each log.

we have 2, 4, 8, 16, 32

Well, you can notice that each time it doubles in the sense that 2*2 = 4, 4 * 2 = 8 ... So in that case, the nth term will be the term before it times two. That means term n will be 2 * (term n-1) and term n-1 is 2 * term n-2, so it keeps going until you get to two. This means it will be the first term times 2*2*2*2*2 n-1 times because we aren't counting the first term. So that is 2*2^n-1 = 2^n.

So basically, it is 2^n for the base of the nth term. usually, when you go through problems like this, you just quickly jump to that conclusion instead of go through all of that logic. So back to the problem.

This means the nth term will be log(base 2^n)8

However, this can be simplified further. From now on I'll just write log instead of log(2^n) for convenience sake. So log 8 = log 2^3 = 3log2

now log(base 2^n)2 = 1/n since 2 is the nth root. So 3 times that is 3/n.

So the nth term is 3/n.

I hope this helped,

Write back if anything wasn't clear,


Robi