Let's solve the problem described in the image. Here is the step-by-step process:

a. Find the exponential growth function

The exponential growth model is given as: $A(t) = A_0 e^{kt}$ where:

• $A_0 = 6.07$ (population in millions in the year 2000),
• $A(t) = 14$ (population in millions in the year 2084),
• $t = 84$ (since 2084 - 2000 = 84 years).

We need to find $k$, the growth rate.

Step 1: Use the formula with given data

$14 = 6.07 e^{k \cdot 84}$

Step 2: Solve for $k$

Divide both sides by $6.07$: $\frac{14}{6.07} = e^{84k}$ $2.305 \approx e^{84k}$

Take the natural logarithm of both sides: $\ln(2.305) = 84k$ $k = \frac{\ln(2.305)}{84}$

Using a calculator: $\ln(2.305) \approx 0.8347$ $k \approx \frac{0.8347}{84} \approx 0.00993$

So, the growth rate is approximately: $k \approx 0.00993$

The exponential growth function:

$A(t) = 6.07 e^{0.00993t}$

b. Find the year when the population reaches 12 million

We now solve for $t$ when $A(t) = 12$: $12 = 6.07 e^{0.00993t}$

Step 1: Solve for $e^{0.00993t}$

Divide both sides by $6.07$: $\frac{12}{6.07} = e^{0.00993t}$ $1.977 \approx e^{0.00993t}$

Step 2: Take the natural logarithm

$\ln(1.977) = 0.00993t$ $t = \frac{\ln(1.977)}{0.00993}$

Using a calculator: $\ln(1.977) \approx 0.683$ $t \approx \frac{0.683}{0.00993} \approx 68.8$

Step 3: Add $t$ to the base year (2000)

$2000 + 68.8 \approx 2069$

Final Answers:

a) The exponential growth function is: $A(t) = 6.07 e^{0.00993t}$

b) The population will reach 12 million in the year 2069.

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

1. How can you graph this exponential function?
2. What happens to the population after 2084 according to the model?
3. How does changing the value of $A_0$ or $k$ affect the growth model?
4. What are the assumptions made in this exponential model?
5. Can the model predict a maximum population limit for the country?

Tip: Always double-check exponential models for accuracy by substituting known values to verify the function.