Question: An ichthyologist observes that the population of a rare fish species in the Amazon doubles every 3 years. If the current population is 324 and it started at a power of 3, what is the smallest exponent \( k \) such that the population reaches at least \( 3^k \)? - Londonproperty
Title: How a Rare Amazon Fish Population Doubles Every 3 Years – Solving for Required Exponent \( k \)
Title: How a Rare Amazon Fish Population Doubles Every 3 Years – Solving for Required Exponent \( k \)
When studying rare species in one of Earth’s most biodiverse regions, ichthyologists often uncover fascinating growth patterns. One such case involves a rare Amazonian fish species whose population doubles every 3 years. Recently, researchers observed that this species currently numbers 324 individuals in the wild — a figure stemming from an initial population expressed as a power of 3. This article explores how to determine the smallest exponent \( k \) such that the population reaches at least \( 3^k \), based on doubling dynamics and exponential growth.
The Exponential Growth Behind the Population
Understanding the Context
The key insight begins with understanding the population’s growth history. We are told:
- The current population is 324.
- The population doubles every 3 years.
- This started from an initial value of \( 3^k \) (but what is \( k \)?).
Our goal is to find the smallest integer exponent \( k \) such that the current population \( 324 \) satisfies:
\[
324 \geq 3^k
\]
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Key Insights
But more challengingly, we want to connect this observed population to its doubling pattern rooted in a power of 3.
Step 1: Express the Current Population in Exponential Form
We start by analyzing the number 324. Factor 324:
\[
324 = 2^2 \ imes 3^4
\]
This shows 324 is not a pure power of 3, but it contains \( 3^4 \) as a factor. This helps contextualize the ichthyologist’s observation: although the real-world population is \( 2^2 \ imes 3^4 \), the growth pattern reflects a doubling every 3 years, consistent with geometric progression driven by exponential scaling.
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Even though 324 is technically \( 3^4 \ imes 4 \), the doubling behavior implies the population grew multiplicatively in powers resembling \( 3^{\cdot} \). To answer the core question — what is the smallest \( k \) such that \( 3^k \leq 324 \)? — we must determine the largest power of 3 less than or equal to 324.
Step 2: Find the Largest \( k \) Such That \( 3^k \leq 324 \)
We compute successive powers of 3:
- \( 3^0 = 1 \)
- \( 3^1 = 3 \)
- \( 3^2 = 9 \)
- \( 3^3 = 27 \)
- \( 3^4 = 81 \)
- \( 3^5 = 243 \)
- \( 3^6 = 729 \)
Now compare:
- \( 3^5 = 243 \leq 324 \)
- \( 3^6 = 729 > 324 \)
Thus, the largest integer \( k \) satisfying \( 3^k \leq 324 \) is \( k = 5 \). This means the population reaches at least \( 3^5 \) — specifically, 243 — with room for future growth.
Step 3: Connect to the Doubling Pattern
Though 324 is not a power of 3, the doubling process every 3 years suggests exponential growth of the form \( P(t) = P_0 \ imes 2^{t/3} \). Suppose after \( t \) years:
\[
P(t) = 3^4 \ imes 4 \ imes 2^{t/3} = 324 \ imes 2^{t/3}
\]
At current observation, \( P(t) = 324 \), and since \( 3^5 = 243 \leq 324 < 729 = 3^6 \), the observed population lies strictly between \( 3^5 \) and \( 3^6 \). Therefore, the smallest exponent \( k \) for which \( 3^k \leq 324 \) is clearly:
