To determine how long it will take Sydney's investment of $440 to grow to $640 with a daily compounding interest rate of 5.6%, we'll use the formula for compound interest when the compounding is not annual:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after \( n \) years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (in decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested for, in years.
We need to solve for \( t \).
Given:
- \( P = $440 \)
- \( A = $640 \)
- \( r = 5.6 \% = 0.056 \) (in decimal form)
- \( n = 365 \) (compounded daily)
Now we want to find \( t \). First, we'll re-arrange the formula to solve for \( t \):
\[ \left(1 + \frac{r}{n}\right)^{nt} = \frac{A}{P} \]
Taking the logarithm of both sides, we have:
\[ nt \cdot \ln\left(1 + \frac{r}{n}\right) = \ln\left(\frac{A}{P}\right) \]
Now solve for \( t \):
\[ t = \frac{\ln\left(\frac{A}{P}\right)}{n \cdot \ln\left(1 + \frac{r}{n}\right)} \]
Plug in the values:
\[ t = \frac{\ln\left(\frac{640}{440}\right)}{365 \cdot \ln\left(1 + \frac{0.056}{365}\right)} \]
Now calculate:
\[ t \approx \frac{\ln\left(\frac{640}{440}\right)}{365 \cdot \ln\left(1 + \frac{0.056}{365}\right)} \]
\[ t \approx \frac{\ln(1.45454545454)}{365 \cdot \ln(1.00015342466)} \]
\[ t \approx \frac{0.374693449}{365 \cdot 0.000153372} \]
\[ t \approx \frac{0.374693449}{0.055979328} \]
\[ t \approx 6.69274580345 \]
Since we want to know the time to the nearest year, we round this to:
\[ t \approx 7 \]
Hence, to the nearest year, it will take Sydney approximately 7 years for the investment to grow from $440 to $640 with a daily compounding interest rate of 5.6%.
