To determine the population of a species that doubles every nine years under ideal conditions, we use the exponential growth model:

$P(t) = P_0 \cdot 2^{\frac{t}{T}}$

Where:

• $P_0$ is the initial population.
• $T$ is the doubling time (in years).
• $t$ is the time (in years) since the start.
• $2^{\frac{t}{T}}$ represents the growth factor based on doubling every $T$ years.

Step-by-step solution:

1. Initial Population ($P_0$): The starting population is $100$.
2. Doubling Time ($T$): The population doubles every $9$ years.
3. Time ($t$): Time elapsed is $t$ years.

The formula becomes: $P(t) = 100 \cdot 2^{\frac{t}{9}}$

Thus, the expression that gives the population of the species $t$ years after the start is:

$P(t) = 100 \cdot 2^{\frac{t}{9}}$

Would you like a graph or numerical examples to understand how the population changes over time?

Related Questions:

1. How would the formula change if the doubling time were 5 years instead of 9 years?
2. What is the population after 18 years under these conditions?
3. How do you derive the doubling formula from basic principles?
4. What is the growth rate per year as a percentage under these conditions?
5. Can this model handle situations where the population cannot grow indefinitely?

Tip:

For exponential growth problems, always identify the doubling (or halving) period and the initial amount first, as these are crucial for constructing the model.