# Geometric Sequences

1. May 14, 2012

### thornluke

1. The problem statement, all variables and given/known data
Find the first term in this geometric sequence that exceeds 500.
2, 4, 8, 16, ...

2. Relevant equations
Un = arn-1

3. The attempt at a solution
a = 2, r = 2
Un = 2 x 2n-1 > 500
2 x (2n)(2-1) > 500
log22 x log22n + log22-1 > log2500
1 x n + (-1) > log2500
n - 1 > log2500
n > log2500 + log22
n > log21000
n > 9.96

But this would make n = 10, whereas the answer should be n =9. What did I do wrong?!

2. May 14, 2012

### Infinitum

The log needs to be taken for the whole expression in the LHS. Why separate the LHS into multiple terms the first place??

3. May 14, 2012

### thornluke

The separation doesn't affect the calculation, it is the same.
log22 x (n-1)log22 > log2500
1 x (n-1) > log2500
n - 1 > log2500
n > log2500 + log22
n > 9.96

4. May 14, 2012

### Infinitum

No no!

I said you need to take the log for the whole LHS. Log is a function and just like, say sin, its taken for the whole expression, rather than each bit. Think of it this way,

$log(2).log(2^{n-1})$

and

$log(2^{n})$

Are they both the same?

Last edited: May 14, 2012
5. May 14, 2012

### HallsofIvy

Staff Emeritus
ln(2(2n-1))= ln(2)+ln(2n-1= 1+ n- 1= n.
Or, more simply, 2(2n-1)= 21+n-1= 2n.

That first "x" should be "+"