# Thread: Negative logarithm while finding limiting difference

1. ## Negative logarithm while finding limiting difference

Hi Guys,

I hit a dead end while solving this problem. I haven't done Complex numbers so far, so I am not sure how to proceed.

By using logarithms, find approximately the difference between $\dfrac{a}{1 - r}$ and $\dfrac{a(1 - r^n)}{1 - r}$ where,

$a = 2, r = -0.7, n = 1001$

I used the difference formula,
$
D = \dfrac{ar^n}{r - 1}
$

and substituting for a, r and n, I got,

$D = \dfrac{2(-0.7)^{1001}}{1.7}$

But for the -0.7 I could have used the logarithm identities to solve this. But I can't take $log -0.7$. The required answer is $-1.032 \times 10^{-155}$

How do I proceed? Thanks again for your help!

Edit:
I went back and calculated $S$ and $S_{\infty}$ individually, They are both identical $=1.176$. Hence the difference would be 0. Is the textbook answer wrong?

2. Hello,

You are asked to compute $D-\frac{a}{r-1}$, not D alone.
At the moment, I can't see why you're asked to use the logarithm... Anyway, $(-0.7)^{1001}=-(0.7)^{1001}$, if it can be of any help...

3. Originally Posted by Moo
Hello,

You are asked to compute $D-\frac{a}{r-1}$, not D alone.
Sorry @Moo, I typed that in word form incorrectly, the question was difference of

$\dfrac{a}{1 - r}$ and $\dfrac{a(1 - r^n)}{1 - r}$

I've edited the question, It still has the negative logarithm.