# exponential function, I can't solve this problem.

• Oct 11th 2009, 07:40 PM
melvis
exponential function, I can't solve this problem.
I've been struggling with this and I even asked 2 friends of mine to help me but they couldn't solve it.
http://www.mathhelpforum.com/math-he...1&d=1255315137
Unfortunately, I can't find the function :Z. I hope somebody can help me. thx 4 u time.
• Oct 11th 2009, 07:51 PM
CFem
An exponential function means this:

$a^x$

If you notice, from 1->2 and 2->3 the value doubles. That's not always a valid hint, but it works for this problem.

So use 2 as the base and see if you can figure out the exponent.
• Oct 11th 2009, 07:53 PM
mr fantastic
Get new friends (Rofl)

And try using $G(n) = \sum_{i=1}^{n} 2^i$.
Quote:

Originally Posted by melvis
I've been struggling with this and I even asked 2 friends of mine to help me but they couldn't solve it.
http://www.mathhelpforum.com/math-he...1&d=1255315137
Unfortunately, I can't find the function :Z. I hope somebody can help me. thx 4 u time.

• Oct 11th 2009, 07:59 PM
TheEmptySet
Quote:

Originally Posted by melvis
I've been struggling with this and I even asked 2 friends of mine to help me but they couldn't solve it.
http://www.mathhelpforum.com/math-he...1&d=1255315137
Unfortunately, I can't find the function :Z. I hope somebody can help me. thx 4 u time.

The pattern that you are looking for is this

$\{2,4,8,16,.... \}=\{2^1,2^2,2^3,...,2^{64} \}$.

So write this as a sum you get $\sum_{n=1}^{64}2^n$

$\sum_{n=0}^{64}2^n=1+\sum_{n=1}^{64}2^n=\frac{1-2^{64+1}}{1-2}=2^{65}-1$

so we get

$1+\sum_{n=1}^{64}2^n=2^{65}-1 \iff \sum_{n=1}^{64}2^n=2^{65}-2=36893488147419103230$

This is alot of rice
• Oct 12th 2009, 12:11 PM
melvis
thx for the help.