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^(n1)
Anywho here is the crux dear reader....
I could write 1/6561 = 1/3 * (1/3)^(n1)
(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(n1)log 1/3
log 1/6561 = log 1/3 + (1n)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)^(n1)
this time I multiply through by 1 and change the power around as
follows:
1/6561 = 1/3 * (1/3)^(1n)
log 1/6561 = 1.log 1/3 + (1n)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(n1)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.