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

Understanding the Context

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.

Key Insights

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
]

Final Thoughts

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.