Solution
For annual compounding, the amount after \(t\) years is
\[ A = P(1 + r)^t \]After 1 year: \[ A = 2000(1.03)^1 = 2060 \] After 2 years: \[ A = 2000(1.03)^2 = 2121.80 \] After 3 years: \[ A = 2000(1.03)^3 = 2185.45 \]
Solution
Simple interest: \[ A = P(1 + rt) = 1000(1 + 0.03 \times 3) = 1090 \] Compound interest: \[ A = 1000(1.03)^3 = 1092.73 \]
The compounded account yields a higher return.
Solution
Simple interest: \[ A = 100(1 + 0.05t) \] Compound interest: \[ A = 100(1.05)^t \]
The compound interest doubles in about 14 years, while the simple interest doubles in about 20 years.
Solution
When interest is compounded \(n\) times per year for \(t\) years, the amount is given by
\[ A = P\left(1 + \frac{r}{n}\right)^{nt} \]Quarterly compounding gives \(n = 4\).
\[ A = 3000\left(1 + \frac{0.05}{4}\right)^{4 \times 5} = 3846.11 \]Solution
First account (4% annually for 2 years):
\[ A = 1200(1.04)^2 = 1297.92 \]Second account (5% annually for 4 years):
\[ A = 1297.92(1.05)^4 = 1577.63 \]Solution
Daily compounding assumes 365 days per year.
First account (4% for 2 years):
\[ A = 1200\left(1 + \frac{0.04}{365}\right)^{365 \times 2} = 1299.94 \]Second account (5% for 4 years):
\[ A = 1299.94\left(1 + \frac{0.05}{365}\right)^{365 \times 4} = 1587.73 \]The final balance is higher than with annual compounding.
Solution
For continuous compounding, the amount after \(t\) years is
\[ A = Pe^{rt} \]First account (4% for 2 years):
\[ A = 1200e^{0.04 \times 2} = 1299.94 \]Second account (5% for 4 years):
\[ A = 1299.94e^{0.05 \times 4} = 1587.75 \]This is essentially the same as daily compounding.
Solution
We are given the final amount \(A = 10{,}000\).
\[ 10{,}000 = P\left(1 + \frac{0.045}{12}\right)^{12 \times 8} \]Solving for \(P\):
\[ P = \frac{10{,}000}{\left(1 + \frac{0.045}{12}\right)^{96}} = 6981.46 \]Solution
Continuous compounding:
\[ A_1 = 120e^{0.07t} \]Annual compounding:
\[ A_2 = 150(1.05)^t \]Set the two amounts equal:
\[ 120e^{0.07t} = 150(1.05)^t \]Taking natural logarithms:
\[ \ln(120) + 0.07t = \ln(150) + t\ln(1.05) \]Solving for \(t\):
\[ t = \frac{\ln(150) - \ln(120)}{0.07 - \ln(1.05)} \approx 10.5 \text{ years} \]
Solution
Continuous compounding:
\[ A_1 = 100e^{0.05t} \]Annual compounding:
\[ A_2 = 100(1.05)^t \]From the graphs, continuous compounding produces a higher balance in the long term.
Solution
We have:
\[ 4000(1 + r)^{10} = 4500 \]Solving for \(r\):
\[ (1 + r)^{10} = \frac{4500}{4000} \] \[ r = e^{\frac{1}{10}\ln(4500/4000)} - 1 \approx 0.012 \]Solution
Balances:
\[ A_1 = 1000e^{0.02t}, \quad A_2 = 500e^{0.08t} \]When balances are equal:
\[ 1000e^{0.02t} = 500e^{0.08t} \] \[ t = \frac{\ln 2}{0.06} \approx 11.5 \text{ years} \]When \(A_2 = 1.5A_1\):
\[ 500e^{0.08t} = 1500e^{0.02t} \] \[ t = \frac{\ln 3}{0.06} \approx 18.5 \text{ years} \]Solution
Each deposit grows for a different length of time.
\[ A_1 = 200\left(1 + \frac{0.04}{4}\right)^4 = 208.12 \] \[ A_2 = 200\left(1 + \frac{0.04}{4}\right)^3 = 206.06 \] \[ A_3 = 200\left(1 + \frac{0.04}{4}\right)^2 = 204.02 \] \[ A_4 = 200\left(1 + \frac{0.04}{4}\right) = 202.00 \] \[ \text{Total} = 820.20 \]Solution
The investment grows continuously at different rates.
\[ A = 1500e^{(0.02 \times 2 + 0.05 \times 1 + 0.06 \times 2)} \] \[ A = 1850.51 \]