A certain type of fluorescent dye used in scientific research loses its brightness over time when exposed to light. Initially, the brightness is at \( 100 \% \) of its original intensity. Each hour, the brightness decreases to \( 80 \% \) of the intensity of the previous hour. The recursive formula describing this decrease is \( B(n)=B(n-1) \cdot 0.80 \), where \( B(n) \) is the brightness as a percentage of the original intensity after \( n \) hours, and \( B(0)=100 \). Determine the number of hours it will take for the dye to reach \( 10 \% \) of its original brightness. (1 point)

To find when the dye reaches \( 10\% \) of its original brightness, we can use the recursive formula. We start with \( B(0) = 100 \). For each hour, brightness becomes \( 80\% \) of the previous hour's brightness. Let's calculate it step-by-step: - \( B(1) = B(0) \cdot 0.80 = 100 \cdot 0.80 = 80 \% \) - \( B(2) = B(1) \cdot 0.80 = 80 \cdot 0.80 = 64 \% \) - \( B(3) = B(2) \cdot 0.80 = 64 \cdot 0.80 = 51.2 \% \) - \( B(4) = B(3) \cdot 0.80 = 51.2 \cdot 0.80 = 40.96 \% \) - \( B(5) = B(4) \cdot 0.80 = 40.96 \cdot 0.80 = 32.768 \% \) - \( B(6) = B(5) \cdot 0.80 = 32.768 \cdot 0.80 = 26.2144 \% \) - \( B(7) = B(6) \cdot 0.80 = 26.2144 \cdot 0.80 = 20.97152 \% \) - \( B(8) = B(7) \cdot 0.80 = 20.97152 \cdot 0.80 = 16.577216 \% \) - \( B(9) = B(8) \cdot 0.80 = 16.577216 \cdot 0.80 = 13.2617728 \% \) - \( B(10) = B(9) \cdot 0.80 = 13.2617728 \cdot 0.80 = 10.60941824 \% \) - \( B(11) = B(10) \cdot 0.80 = 10.60941824 \cdot 0.80 = 8.487534592 \% \) After calculating for hours, we see that at \( n = 10 \), the brightness reaches approximately \( 10.6\% \), and by \( n = 11 \), it drops below \( 10\% \). Therefore, it will take \( 10 \) hours for the dye to first reach approximately \( 10\% \) of its original brightness.