Space Station Population Capacity
A space station on a recolonization mission has a population that quadruples every 54 years, and a life support system whose capacity only doubles every 54 years. Initially, the population is 2 people and the life support capacity is 16,384 people.
How many years will pass before the population first reaches (equals) the capacity of the life support system?
Hints
- Both quantities grow exponentially with the same 54-year period. Express each as a function of $k$, the number of 54-year periods, rather than time in years.
- Write $P(k) = 2 \cdot 4^k$ and $C(k) = 16384 \cdot 2^k$. The population grows faster (base 4 vs base 2), but starts much smaller -- set them equal and solve for $k$.
- Convert $4^k = 2^{2k}$ so everything is in powers of 2. Then $2^{2k+1} = 2^{k+14}$ gives $k = 13$, corresponding to $t = 702$ years.
Worked Solution
How to Think About It: Both quantities grow exponentially with the same period of 54 years, but different base rates: population grows as $4^k$ and capacity grows as $2^k$, where $k$ counts 54-year periods. Population grows faster (base 4 vs. base 2), but it starts far behind (2 vs. 16,384). The question is just asking when the faster-growing quantity catches up. A quick order-of-magnitude check: $4^k / 2^k = 2^k$, so the ratio of population to capacity grows as $2^k / 8192$ -- it starts tiny and doubles every period. We need $2^k / 8192 = 1$, meaning $2^k = 8192 = 2^{13}$, so $k = 13$. That is $13 \times 54 = 702$ years.
Quick Estimate: At $k = 10$, population is $2 \cdot 4^{10} \approx 2 \times 10^6$ and capacity is $16384 \cdot 2^{10} \approx 1.7 \times 10^7$. Population is still about 8x smaller. At $k = 13$, population is $2 \cdot 4^{13} \approx 1.34 \times 10^8$ and capacity is $16384 \cdot 2^{13} \approx 1.34 \times 10^8$. They match. Confirmed.
Formal Solution:
Let $k = t / 54$ denote the number of 54-year periods elapsed. Then: $$P(k) = 2 \cdot 4^k \qquad \text{and} \qquad C(k) = 16384 \cdot 2^k$$
Set population equal to capacity: $$2 \cdot 4^k = 16384 \cdot 2^k$$
Since $4^k = (2^2)^k = 2^{2k}$: $$2 \cdot 2^{2k} = 16384 \cdot 2^k$$ $$2^{2k+1} = 2^{14} \cdot 2^k$$ $$2^{2k+1} = 2^{k+14}$$
Equating exponents: $$2k + 1 = k + 14 \implies k = 13$$
Converting back to years: $$t = 54k = 54 \times 13 = 702 \text{ years}$$
Answer: The population reaches the life support capacity after $\boxed{702}$ years.
Intuition
This is a classic exponential race -- two quantities growing at different rates starting from different initial values. The key insight is to work in the natural unit of the problem (54-year periods rather than years) and express both quantities as powers of 2. Once you do that, the algebra is just matching exponents.
The deeper lesson is about the power of multiplicative compounding: the population starts at 2 while capacity is at 16,384 -- a factor of 8,192 behind. Yet because population doubles its gap-closing ratio every period, it only takes 13 periods to catch up. In finance, this same structure appears in leverage and compounding return comparisons: a fund with a higher compounding rate will always eventually overtake a larger starting balance, given enough time.