Solution: The population grows as \( 324 \times 2^t/3 \), where \( t \) is time in years. We want the smallest \( k \) such that: - Treasure Valley Movers

Title: Understanding Population Growth: The Case of ( 324 \ imes 2^{t/3} )

Introduction
Population growth is a key topic in demography, urban planning, and environmental studies. One population model describes growth using the formula ( P(t) = 324 \ imes 2^{t/3} ), where ( P(t) ) represents the population at time ( t ) in years, and ( 324 ) is the initial population. This exponential model reveals how quickly populations expand—especially under favorable conditions. In this article, we explore the meaning of this function and highlight the smallest integer ( k ) such that population levels remain valid and meaningful over time.

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The Mathematics Behind Population Growth

The given population function is:
[
P(t) = 324 \ imes 2^{t/3}
]
Here:
- ( 324 ) is the initial population (at ( t = 0 )),
- ( t ) is time in years,
- The base ( 2 ) and exponent ( t/3 ) indicate exponential growth with a doubling time of 3 years.

This means every 3 years, the population doubles:
[
P(3) = 324 \ imes 2^{3/3} = 324 \ imes 2 = 648 <br/>P(6) = 324 \ imes 2^{6/3} = 324 \ imes 4 = 1296 <br/>\ ext{and so on.}
]

Exponential growth like this approximates real-world population dynamics in many developing regions, especially when resources remain abundant and external factors do not abruptly alter growth trends.

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When Does This Model Remain Valid?

While the formula ( P(t) = 324 \ imes 2^{t/3} ) is powerful for prediction, practical application requires considering when the population becomes meaningful or reaches physical, environmental, or societal thresholds. The smallest ( k ) would represent the earliest time step—years—where the population becomes substantial enough to trigger meaningful impact, policy response, or data threshold reaching.

Let’s define a meaningful marker: suppose the smallest sustainable or observable population for analysis or planning begins at ( P(k) \geq 1000 ). We solve for the smallest integer ( k ) such that:
[
324 \ imes 2^{k/3} \geq 1000
]

Solve the Inequality:
[
2^{k/3} \geq rac{1000}{324} pprox 3.086
]

Take logarithm base 2 of both sides:
[
rac{k}{3} \geq \log_2(3.086)
]

Using ( \log_2(3.086) pprox \log_2(3.09) pprox 1.63 ) (since ( 2^{1.63} pprox 3.09 )),
[
rac{k}{3} \geq 1.63 \quad \Rightarrow \quad k \geq 4.89
]

So the smallest integer ( k ) is ( 5 ).

Verify:
- ( k = 4 ): ( P(4) = 324 \ imes 2^{4/3} pprox 324 \ imes 2.5198 pprox 816 )
- ( k = 5 ): ( P(5) = 324 \ imes 2^{5/3} pprox 324 \ imes 3.1748 pprox 1027 \geq 1000 )

Hence, ( k = 5 ) is the smallest integer time at which the population crosses the 1000 threshold.

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Why ( k = 5 ) Matters in Policy and Planning

Identifying ( k = 5 ) helps stakeholders in healthcare, education, housing, and environmental management anticipate demand shifts:
- Beyond year 5, population growth accelerates significantly, requiring expanded infrastructure.
- Decision-makers can use this threshold to prepare resources before rapid expansion begins.
- Analysts tracking long-term trends can use this step to refine models, adjusting for resource limits or urban planning constraints.

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Conclusion