A quantity with an initial value of 8100 grows exponentially at a rate of 0.35% every 4 decades. What is the value of the quantity after 47 years, to the nearest hundredth?

Answer 1

To solve this problem, we first need to find the growth factor for the quantity. The growth factor is given by the formula:

\[ growth\ factor = 1 + \frac{r}{100} \]

where r is the growth rate. In this case, the growth rate is 0.35%, so:

\[ growth\ factor = 1 + \frac{0.35}{100} = 1.0035 \]

Since the quantity grows every 4 decades, we need to find the number of growth periods in 47 years. Since 1 decade is equal to 10 years, there are \( \frac{47}{10} \) = 4.7 decades in 47 years. This means there have been 5 growth periods (4 decades) up to 40 years, and an additional 1.7 growth periods for the remaining 7 years.

Now, we can calculate the value of the quantity after 47 years using the formula for exponential growth:

\[ final\ value = initial\ value \times (growth\ factor)^{number\ of\ growth\ periods} \]

Plugging in the values, we get:

\[ final\ value = 8100 \times (1.0035)^5 \times (1.0035)^{0.7} \]

\[ final\ value ≈ 8100 \times 1.017675 \times 1.007676 \]

\[ final\ value ≈ 8100 \times 1.025692 \]

\[ final\ value ≈ 8312.30 \]

Therefore, the value of the quantity after 47 years is approximately \( $8312.30 \) to the nearest hundredth.