Sum of a sequence that has an answer, but I can't get that answer by hand Find n if
$a_1 = 4$
$r = -4$
$S_n = 13108$
Using the formula
$S_n = \frac{a_1(1 - (r)^n)}{1 - r}$ I would have
$13108 = \frac{4(1 - (-4)^n)}{5}$
$65540 = 4(1 - (-4)^n)$
$16385 = 1 - (-4)^n$
$16384 = -(-4)^n$
$-16384 = (-4)^n$
I can't take the log now though because I have negatives! I know the answer is 7 because if I plug in n = 7, I do get the desired result. However, I'm not sure how to get this by hand. Can someone please clear this up?
 A: $$\begin{split}-16384&=(-4)^n\\
16384&=(-1)^{n-1}4^n \text{     (equation 2)}
\end{split}$$
At this point there will be a solution if for the $n$ such that $4^n=16384$, we have that $(-1)^{n-1}$ is just $1$. In terms of integer solutions, there would be a solution if $4^n=16384$ for some odd $n$. We focus on $n=\frac{\log 16384}{\log 4}=7$, and verify that this indeed works in the original equation as $(-1)^{7-1}=1$ so the lhs and rhs are equal in equation 2. If instead $n$ were some even integer such that $4^n=16384$, there would be no solution due to the sign mismatch.
A: We know that $-16384 = (-4)^n$. Taking the absolute values of both sides, we conclude that $16384 = 4^n$. We then take $\log_4$ of both sides to get that $n = 7$.
We must check that $n = 7$ actually gives us the correct value of $S_n$ by plugging $n$ back into the formula (otherwise, there could be no solutions). Thankfully, it does.
A: $$-16348=(-4)^n \implies n=2p+1$$
$$\implies -16384=-(-4)^{2p}$$
$$\implies 16384=(16)^p$$
$$p=\frac{\ln(16384)}{\ln(16)}=3$$
$$\implies n=7$$
