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(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.

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)^n1 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)^(n1)
log 1/1594323 = log 1/3 +(n1)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)^(n1)
log 1/243 = log 1/3 + (n1)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)^(n1)
log 567/15625 = log 7/9 + (n1)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