## Algebra 2 (1st Edition)

$n \gt 7.044$
From the previous part it can be seen that $a_1=66; r=\dfrac{1}{2}$. and $a_n=66 (\dfrac{1}{2})^{n-1}$ We need the n-th term to be less than 1 gram. Thus, $a_n \lt 1 \implies 66 (\dfrac{1}{2})^{n-1} \lt 1$ or, $(n-1) \log (\dfrac{1}{2}) \lt \log (\dfrac{1}{66})$ We will divide the inequality with $\log (1/2)$, which is negative. Thus, we have $n-1 \gt \dfrac{\log (1/66)}{\log (1/2)}$ or, $n \gt 7.044$