Geometric Sequences and Series

**GrigOrig99** · July 24th 2011, 10:41 AM

Q. If Un of the sequence 2, 4, 8,... is 512, find n.

Eq.: Un = ar^n-1

Attempt:
Un = ar^n-1
2.2^n-1 = 512
n = 9

My question is, can n be solved without randomly subbing values into n? Thank you.

**waqarhaider** · July 24th 2011, 10:50 AM

you can use logarithms

**skeeter** · July 24th 2011, 11:06 AM

Originally Posted by GrigOrig99

Q. If Un of the sequence 2, 4, 8,... is 512, find n.

Eq.: Un = ar^n-1

Attempt:
Un = ar^n-1
2.2^n-1 = 512
n = 9

My question is, can n be solved without randomly subbing values into n? Thank you.

the author of these "formulated" problems expects that the student knows

$512 = 2^9$

**Siron** · July 24th 2011, 11:50 AM

Originally Posted by waqarhaider

[Tex]\mbox{you\can\use\logarithms[\Tex]

I don't think GrigOrig99 has already learned logarithms.

**Archie Meade** · July 24th 2011, 11:55 AM

Originally Posted by GrigOrig99

Q. If Un of the sequence 2, 4, 8,... is 512, find n.

Eq.: Un = ar^n-1

Attempt:
Un = ar^n-1
2.2^n-1 = 512
n = 9

My question is, can n be solved without randomly subbing values into n? Thank you.

$u_1=2=2^1$

$u_2=4=2^2$

$u_3=8=2^3$

$u_n=2^n$

If you wish, you may keep doubling until you reach 512

$u_4=16,\;\;\;u_5=32,\;\;\;u_6=64....\;\;etc$

**GrigOrig99** · July 24th 2011, 12:19 PM

Apologies if I was unclear. I understand why n = 9, i.e. 2^9 = 512. What I meant was whether or not 9, itself, can be calculated from just:

2.2^n-1 = 512

I don't believe it's possible but just wanted to double check with you guys. Thanks again.

**Archie Meade** · July 24th 2011, 12:39 PM

You may use the Power Law of logarithms.

$2^n=512\Rightarrow\ log2^n=log512$

Now use

$loga^b=bloga$

$nlog2=log512$

$n=\frac{log512}{log2}$

Natural logs or "logs to base 10" may be used.

**GrigOrig99** · July 24th 2011, 01:14 PM

Thank you for clearing that up. I'm getting better at doing most of these sums on my own, slowly but surely.

**Siron** · July 24th 2011, 01:34 PM

Sorry, then I've to take back what I said about that you haven't learnt logarithms yet

.

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