Let's solve the problem step by step:

(A) Exponential Model Form

The general form of the exponential model is:

$P = a \cdot b^t$

Where:

• $a$ is the initial population.
• $b$ is the growth factor.

1. Initial Population ($a$): The population starts at 18,000. Therefore: $a = 18000$

2. Growth Factor ($b$): The population grows by 2.3% per year. This means the population increases by a factor of $1 + 0.023 = 1.023$ each year. Hence: $b = 1.023$

Thus, the equation becomes: $P = 18000 \cdot 1.023^t$

(B) Using Logs to Determine the Number of Years

We need to determine how long it takes for the population to reach 79,200. Substituting $P = 79200$ into the equation:

$79200 = 18000 \cdot 1.023^t$

1. Divide both sides by 18,000: $\frac{79200}{18000} = 1.023^t$ $4.4 = 1.023^t$

2. Take the natural logarithm ($\ln$) of both sides: $\ln(4.4) = \ln(1.023^t)$

3. Apply the logarithm power rule ($\ln(a^b) = b \ln(a)$): $\ln(4.4) = t \cdot \ln(1.023)$

4. Solve for $t$: $t = \frac{\ln(4.4)}{\ln(1.023)}$

Now, calculate using logarithms:

• $\ln(4.4) \approx 1.4816$
• $\ln(1.023) \approx 0.0227$

$t = \frac{1.4816}{0.0227} \approx 65.25$

Thus, it will take approximately 65.25 years for the population to reach 79,200.

Final Answers:

(A) Exponential Model: $P = 18000 \cdot 1.023^t$

(B) Time to reach 79,200 organisms: $t = 65.25 \, \text{years}$

Would you like further clarification or details? Here are some related questions to explore:

1. What happens to the population if the growth rate changes to 5%?
2. How would the equation change if the initial population were doubled?
3. Can we use a different logarithm base (e.g., base 10) to solve part (B)?
4. How long would it take for the population to double?
5. What does the exponential growth model imply about population trends in the long term?

Tip: When solving exponential equations, taking the logarithm is a powerful method to isolate the variable in the exponent.