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Let's break down the problems one by one.

### Question 7:
The house prices in Hobart are expected to grow by 8% per year for the next 5 years. The current value is $370,000. We are asked to find the expected value in 5 years.

This is a compound growth problem and can be solved using the formula:

\[
A = P \times (1 + r)^t
\]

Where:
- $A$ is the amount after $t$ years
- $P$ is the initial amount (370,000)
- $r$ is the growth rate per year (8% or 0.08)
- $t$ is the number of years (5)

Substitute the values:

\[
A = 370,000 \times (1 + 0.08)^5
\]
\[
A = 370,000 \times (1.08)^5
\]
\[
A \approx 370,000 \times 1.4693
\]
\[
A \approx 543,641
\]

Thus, the expected value of the house in 5 years is approximately **$543,641**.

---

### Question 8:
The population of a country town is expected to fall by 15% per year for the next 8 years. The current population is 22,540. We are asked to find the expected population in 8 years.

This is a case of population decay, using a similar formula:

\[
P_t = P_0 \times (1 - r)^t
\]

Where:
- $P_t$ is the population after $t$ years
- $P_0$ is the initial population (22,540)
- $r$ is the rate of decline per year (15% or 0.15)
- $t$ is the number of years (8)

Substitute the values:

\[
P_t = 22,540 \times (1 - 0.15)^8
\]
\[
P_t = 22,540 \times (0.85)^8
\]
\[
P_t \approx 22,540 \times 0.27249
\]
\[
P_t \approx 6,141
\]

Thus, the expected population in 8 years is approximately **6,141**.

---

### Question 6:
We need to find the initial investment amount (present value) for each given case, assuming compound interest. The formula is:

\[
P = \frac{A}{(1 + r)^t}
\]

Where:
- $P$ is the initial investment
- $A$ is the final amount
- $r$ is the annual interest rate (expressed as a decimal)
- $t$ is the time period in years

Let's calculate for each case:

**(a)** $A = 26,500$, $r = 0.04$, $t = 3$

\[
P = \frac{26,500}{(1 + 0.04)^3} = \frac{26,500}{1.124864} \approx 23,561
\]

**(b)** $A = 42,000$, $r = 0.06$, $t = 4$

\[
P = \frac{42,000}{(1 + 0.06)^4} = \frac{42,000}{1.262477} \approx 33,274
\]

**(c)** $A = 35,500$, $r = 0.035$, $t = 6$

\[
P = \frac{35,500}{(1 + 0.035)^6} = \frac{35,500}{1.229874} \approx 28,865
\]

**(d)** $A = 28,200$, $r = 0.047$, $t = 2$

\[
P = \frac{28,200}{(1 + 0.047)^2} = \frac{28,200}{1.096209} \approx 25,729
\]

---

If you have any further questions or need additional details, feel free to ask!

### Follow-up Questions:
1. How would the calculations change if the growth/decline rates were compounded monthly instead of annually?
2. Can you explain how to derive the compound interest formula for continuous growth?
3. What would happen if the interest rate or growth rate were negative for one of the scenarios?
4. How would you adjust the formula for growth if the rate changed every year?
5. In a real-world context, how do taxes and inflation impact these growth/decay models?

### Tip:
When dealing with growth or decay problems, always ensure you correctly interpret the rate (growth vs. decay) and the compounding period (annually, monthly, etc.) to use the appropriate formula.

Average house prices in Hobart are expected to grow by 8% per year for the next 5 years. What is the expected average value of a house in Hobart in 5 years' time, to the nearest dollar, if it is currently valued at $370,000? Also, calculate the expected population of a country town with a decline of 15% per year for 8 years if the current population is 22,540 people. Additionally, find the initial investment amounts to the nearest dollar given final balances, annual interest rates, and time periods for the following scenarios: (a) $26,500, 4%, 3 years, (b) $42,000, 6%, 4 years, (c) $35,500, 3.5%, 6 years, (d) $28,200, 4.7%, 2 years.

Learn how to calculate exponential growth for house prices in Hobart with an 8% annual increase, expected population decay at 15% over 8 years, and initial investment values for compound interest problems. Includes key concepts such as compound interest and exponential decay.

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