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空间站人口

Space Station

专题
Discrete Math / 离散数学
难度
L2

题目详情

一座执行再殖民任务的空间站依靠可扩展生命支持系统维持乘员生存。站内人口每 54 年变为原来的 4 倍，而生命支持系统容量每 54 年只能翻倍。若初始人口为 2 人，初始容量为 16384 人，多少年后人口会达到生命支持系统的容量上限？

The population of a space station on a recolonization mission is kept alive by its expandable life support system. Given they are trying to repopulate, the population aboard the station quadruples every 54 years, but the life support system capacity can only be doubled every 54 years. Suppose the initial population is 2 people and the initial life support system has a capacity for 16384 people. How many years will pass before the population reaches the capacity of the support system?

解析

设每 54 年为一个周期，记 k=t54k=\frac{t}{54}。

人口函数为 P(k)=2⋅4kP(k)=2\cdot 4^k 生命支持系统容量为 C(k)=16384⋅2kC(k)=16384\cdot 2^k

当人口首次达到容量时，令两者相等：

2⋅4k=16384⋅2k2\cdot 4^k = 16384\cdot 2^k

因为 4k=22k4^k=2^{2k}，所以上式化为

22k+1=2k+142^{2k+1}=2^{k+14}

于是

k=13k=13

换回年数：

t=54k=54⋅13=702 年t = 54k = 54\cdot 13 = \boxed{702\ \text{年}}

Original Explanation

Our goal is to find the time (in years) when the population first reaches the system’s capacity. To do this, we compare the exponential growth of the population and the exponential growth of the capacity.

Step #1: Define the number of growth periods

Because both the population and capacity grow in discrete 54-year cycles, we define:

k=t54k = \frac{t}{54}

where

  • kk is the number of 54-year periods
  • tt is the total number of years

This allows us to write both quantities as exponential functions of kk

Step #2: Write the population as a function of kk

The population starts at 2 and quadruples every period, so:

P(k)=2⋅4kP(k) = 2 \cdot 4^k

In each 54-year interval, the population is multiplied by 4. After kk intervals, this gives 4k4^k. Multiplying by the initial population gives the full expression.

Step #3: Write the life support capacity as a function of kk

The capacity begins at 16,384 and doubles every period, so:

C(k)=16384⋅2kC(k) = 16384 \cdot 2^{k}

Each period multiplies the capacity 2. After kk periods, this produces 2k2^k. Again, multiplying by the initial capacity gives the full capacity after kk growth cycles.

Step #4: Set population equal to capacity and solve for kk

2⋅4k=16384⋅2k4k=(22)k=22k2⋅22k=16384⋅2k16384⋅2k=214⋅2k=2k+1422k+1=2k+142k+1=k+14k=132 \cdot 4^{k} = 16384 \cdot 2^{k} \\ 4^{k} = (2^{2})^{k} = 2^{2k} \\ 2 \cdot 2^{2k} = 16384 \cdot 2^{k} \\ 16384 \cdot 2^{k} = 2^{14} \cdot 2^{k} = 2^{k + 14} \\ 2^{2k + 1} = 2^{k + 14} \\ 2k + 1 = k + 14 \\ k = 13

Step #5: Convert number of periods back to years

Each period is 54 years, so:

t=54k=54⋅13=702 yearst = 54k = 54 \cdot 13 = \boxed{702 \ \text{years}}