Solving the inequality $1+1.2 + 1.2^2 + 1.2 ^3 + \cdots + 1.2^n < N/16?$ Given the inequality $$1+1.2 + 1.2^2 + 1.2 ^3 +\cdots + 1.2^n < \frac{N}{16}?$$
I need the value of $n$, or just an approximation. $N$ is known. 

$1+1.2+1.2^2+\ldots +1.2^n = \frac{1.2^{n+1}-1}{1.2-1}$ $= 5(1.2^{n+1}-1)$.
So you want to find the values of $n$ such that the inequality
$$5(1.2^{n+1}-1) < \frac{N}{16}$$
This gives
$$1.2^{n+1} < \frac{N}{80} - 1$$
Taking logs this gives
$$n \log 1.2 < \log \left(\frac{N}{80}-1\right)$$
which as $\log 1.2 < .2$ and $\log 80 \approx .4.4$ , will hold for large $N$ if the following is satisfied:
$$.2n < \log N - 4.5$$
or $$n < 5 \log N - 18$$
