 Peter's GP Log question - XP Math - Forums Sign Up FREE! | Sign In | Classroom Setup | Common Core Alignment  XP Math - Forums Peter's GP Log question 02-20-2011 #1 Lisasmith111 Join Date: Nov 2010 Posts: 36 Peter's GP Log question Here is something I found by poking around with Peter's Series. Question is why does it work (sort of!). ie 1/3 - 1/9 + 1/27... Suppose I give you a term from it say -1/6561. Now we know the common ratio could be calculated as follows:- term 2 / term 1 = -1/9 / 1/3 = -1/3 so common ratio = -1/3 we also know the first term = 1/3 Which means we could use the formula for finding a given term from the series ar^(n-1) Anywho here is the crux dear reader.... I could write -1/6561 = 1/3 * (-1/3)^(n-1) (Just to be clear in words the RHS is a third times minus a third raised to the power of n minus 1.) If I solve for n using logs I get -1 * log 1/6561 = log 1/3 + -1(n-1)log 1/3 -log 1/6561 = log 1/3 + (1-n)log 1/3 -log 1/6561 = log 1/3 + log 1/3 -nlog 1/3 (-log 1/6561 - 2log 1/3) / -log 1/3 = n = 10 (which is the wrong answer) However, If I do the following I get an answer which is closer to the truth:- So again I write -1/6561 = 1/3 * (-1/3)^(n-1) this time I multiply through by -1 and change the power around as follows:- 1/6561 = -1/3 * (1/3)^(1-n) log 1/6561 = -1.log 1/3 + (1-n)log 1/3 log 1/6561 = -log 1/3 +log 1/3 -nlog 1/3 log 1/6561 = -nlog 1/3 log 1/6561 / log 1/3 = -n n = -8 Now we can't have a negative term. So I could say we want the absolute value of n and therefore we get n = 8. Ie -1/6561 is the 8th term in the series. Which is correct. The trick seems to work but why does is work? (at least sort of!). But why? Is there a better way of doing it?   02-22-2011   #2
MAS1   Join Date: Dec 2008
Posts: 249 Quote:
 Originally Posted by Lisasmith111 Here is something I found by poking around with Peter's Series. Question is why does it work (sort of!). ie 1/3 - 1/9 + 1/27... Suppose I give you a term from it say -1/6561. Now we know the common ratio could be calculated as follows:- term 2 / term 1 = -1/9 / 1/3 = -1/3 so common ratio = -1/3 we also know the first term = 1/3 Which means we could use the formula for finding a given term from the series ar^(n-1) Anywho here is the crux dear reader.... I could write -1/6561 = 1/3 * (-1/3)^(n-1) (Just to be clear in words the RHS is a third times minus a third raised to the power of n minus 1.) If I solve for n using logs I get -1 * log 1/6561 = log 1/3 + -1(n-1)log 1/3 -log 1/6561 = log 1/3 + (1-n)log 1/3 -log 1/6561 = log 1/3 + log 1/3 -nlog 1/3 (-log 1/6561 - 2log 1/3) / -log 1/3 = n = 10 (which is the wrong answer) However, If I do the following I get an answer which is closer to the truth:- So again I write -1/6561 = 1/3 * (-1/3)^(n-1) this time I multiply through by -1 and change the power around as follows:- 1/6561 = -1/3 * (1/3)^(1-n) log 1/6561 = -1.log 1/3 + (1-n)log 1/3 log 1/6561 = -log 1/3 +log 1/3 -nlog 1/3 log 1/6561 = -nlog 1/3 log 1/6561 / log 1/3 = -n n = -8 Now we can't have a negative term. So I could say we want the absolute value of n and therefore we get n = 8. Ie -1/6561 is the 8th term in the series. Which is correct. The trick seems to work but why does is work? (at least sort of!). But why? Is there a better way of doing it?
Hi Lisa:

In your attempt to solve the problem using logs, your log step is incorrect.
(-1/6561) = (1/3)(-1/3)^(n - 1)

Now you have to take the logarithm of both sides which gives:
log(-1/6561) = log[(1/3)(-1/3)^(n - 1)]
NOT -1 * log 1/6561 = log 1/3 + -1(n-1)log 1/3 which is what you have.

log(-1/6561) is not equal to -1*log(1/6561)

As you probably know taking the log of a negative number is not defined, so you cannot use that method.   02-23-2011   #3
Lisasmith111 Join Date: Nov 2010
Posts: 36 GP and Logs

Quote:
 Originally Posted by MAS1 Hi Lisa: In your attempt to solve the problem using logs, your log step is incorrect. (-1/6561) = (1/3)(-1/3)^(n - 1) Now you have to take the logarithm of both sides which gives: log(-1/6561) = log[(1/3)(-1/3)^(n - 1)] NOT -1 * log 1/6561 = log 1/3 + -1(n-1)log 1/3 which is what you have. log(-1/6561) is not equal to -1*log(1/6561) As you probably know taking the log of a negative number is not defined, so you cannot use that method.
Hello Mr Hui, Mas and friends,

I've had a rethink and come up with the following:-

If we ignore the negatives in all the variables to the equation as follows it seems to do the trick.

Example 1

consider Peter's GP.

1/3 - 1/9 + 1/27...

we know the common ratio is -1/3 and the first term is 1/3. Now, suppose we are given the following term from the series -1/1594323 and were asked to find

the number of the term.

term = a(r)^n-1 and solve for n (the number of the term remember!)

so in our question we have

- First term = 1/3 (+ve no problem)

- Common ratio = -1/3 (so take absolute value = 1/3 for our purposes)

- Finally the term that we are trying to find the number of -1/1594323 and again take the absolute value we get + 1/1594323.

Slot them into the eqn above and we get:-

1/1594323 = 1/3 * (1/3)^(n-1)

log 1/1594323 = log 1/3 +(n-1)log 1/3

log 1/1594323 = log 1/3 + nlog 1/3 -log 1/3

so n = (log 1/1594323 - log 1/3 + log 1/3) / log 1/3

n = log 1/1594323 / log 1/3

n = 13

so -1/1594232 is the 13th term in the series.

Example 2

Consider the following GP

-1/3, 1/9, -1/27, 1/81, -1/243, 1/729...

we can calculate the number of the term for -1/243. we know already it is the 5th term but can we find it with logs?

First term = -1/3 becomes 1/3

Common ratio = -1/3 becomes 1/3

Term to find number of = -1/243 becomes 1/243

1/243 = 1/3 * (1/3)^(n-1)

log 1/243 = log 1/3 + (n-1)log 1/3

Log 1/243 = log 1/3 + nlog 1/3 = log 1/3

log 1/243 = nlog 1/3

n = log 1/243 / log 1/3

n = 5

So -1/243 is the 5th term in the series.

Example 3

Consider the following series:-

-7/9, 7/15, -7/25, 21/125...

From the information above and the following term from the series -567/15625 determine the number of the term within the series.

Common ratio = -3/5 becomes 3/5

first term = -7/9 becomes 7/9

term we need number of -567/15625 becomes 567/15625

So we have 567/15625 = 7/9 * (3/5)^(n-1)

log 567/15625 = log 7/9 + (n-1)log 3/5

log 567/15625 = log 7/9 + nlog 3/5 - log 3/5

n = (log 567/15625 - log 7/9 + log 3/5) / log 3/5

n = 7

So -567/15625 is the seventh term in the series.

check:-

-7/9, 7/15, -7/25, 21/125, -63/625, 189/3125, -567/15625

LIsa   02-23-2011 #4 MAS1   Join Date: Dec 2008 Posts: 249 Can you look at example 3 again? I think there is something wrong with it. It doesn't look like a geometric series.   02-23-2011   #5
Lisasmith111 Join Date: Nov 2010
Posts: 36 Quote:
 Originally Posted by MAS1 Can you look at example 3 again? I think there is something wrong with it. It doesn't look like a geometric series.
Hello Mas,

GP from example 3 is derived as follows:-

Common ratio = 7/15 / -7/9 = -3/5

Hence we have the following terms:-

1st term = -7/9

2nd term = -7/9 * -3/5 = 7/15

3rd term = 7/15 * -3/5 = -7/25

4th term = -7/25 * -3/5 = 21/125

5th term = 21/125 * -3/5 = -63/625

6th term = -63/625 * -3/5 = 189/3125

7th term = 189/3125 * -3/5 = -567/15625

Hence -7/9, 7/15, -7/25, 21/125, -63/625, 189/3125, -567/15625...

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