To determine how many days it takes for the bacteria count to double with a daily 5% increase, follow these steps:

Step 1: Model the growth

Let the initial count be (N_0). After (t) days, the count is:
(N(t) = N_0 \times (1.05)^t)

Step 2: Set up the doubling condition

We want (N(t) = 2N_0):
(2N_0 = N_0 \times (1.05)^t)
Cancel (N_0):
(2 = (1.05)^t)

Step 3: Solve for (t)

Take natural logs on both sides:
(\ln(2) = t \times \ln(1.05))
(t = \frac{\ln(2)}{\ln(1.05)} \approx \frac{0.6931}{0.04879} \approx 14.21)

Step 4: Round up to the next day

Since the count isn’t doubled after 14 days (((1.05)^{14} \approx 1.98 < 2)), but doubles by day 15 (((1.05)^{15} \approx 2.08 > 2)), the answer is 15 days.

Answer: 15

（免责声明：本文为本网站出于传播商业信息之目的进行转载发布，不代表本网站的观点及立场。本文所涉文、图、音视频等资料的一切权利和法律责任归材料提供方所有和承担。本网站对此资讯文字、图片等所有信息的真实性不作任何保证或承诺，亦不构成任何购买、投资等建议，据此操作者风险自担。） 本文为转载内容，授权事宜请联系原著作权人，如有侵权，请联系本网进行删除。