Geometric Sequences: Find 1st Term Exceeding 500

[thornluke]

Homework Statement

Find the first term in this geometric sequence that exceeds 500.
2, 4, 8, 16, ...

Homework Equations

U_n = ar^n-1

The Attempt at a Solution

a = 2, r = 2
U_n = 2 x 2^n-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?!

 

[Infinitum]

2 x (2ⁿ)(2^-1) > 500
log₂2 x log₂2ⁿ + log₂2^-1 > log₂500

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

 

[thornluke]

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

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

 

[Infinitum]

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

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?

 

[HallsofIvy]

Homework Statement

Find the first term in this geometric sequence that exceeds 500.
2, 4, 8, 16, ...

Homework Equations

U_n = ar^n-1

The Attempt at a Solution

a = 2, r = 2
U_n = 2 x 2^n-1 > 500
2 x (2ⁿ)(2^-1) > 500

ln(2(2^n-1))= ln(2)+ln(2^n-1= 1+ n- 1= n.
Or, more simply, 2(2^n-1)= 2^1+n-1= 2ⁿ.

log₂2 x log₂2ⁿ + log₂2^-1 > log₂500
1 x n + (-1) > log₂500

That first "x" should be "+"

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?!