Question

Suppose a certain amount of money is invested and draws interest compounded continuously. (a) If the original amount doubles in two years, then what is the annual interest rate? (b) If the original amount increases \(50 \%\) in six months, then how long will it take the original amount to double?

Short answer

(a) The annual interest rate is approximately 34.66%. (b) It takes approximately 2 years for the original amount to double when it increases by 50% in six months.

Step-by-step solution

Part A: Calculate the annual interest rate

We are given that the original amount doubles in two years, therefore A = 2 * P, and t = 2. Using the continuous compound interest formula: \(2P = P * e^{2r}\) Now, we need to solve for r. Step 1: Divide by P on both sides of the equation: \(\frac{2P}{P} = \frac{P * e^{2r}}{P}\) Simplifying, we get: \(2 = e^{2r}\) Step 2: Use the natural logarithm function (ln) to eliminate e from the equation: \(ln(2) = ln(e^{2r})\) Step 3: Use the property of logarithms that allows us to bring the exponent down: \(ln(2) = 2r * ln(e)\) Since \(ln(e) = 1\), we have: \(ln(2) = 2r\) Step 4: Divide by 2 to solve for r: \(r = \frac{ln(2)}{2}\) The annual interest rate is r ≈ 0.3466, or approximately 34.66%.

Part B: Calculate the time to double the original amount

We are given that the original amount increases by 50% in six months, so A = 1.5 * P, t = 0.5 (since six months is equal to 0.5 years). We will use the continuously compounded interest formula again: \(1.5P = P * e^{0.5r}\) Now we need to solve for the time it takes for the original amount to double, keeping in mind that we found the interest rate in part (a). Step 1: Divide by P on both sides of the equation: \(\frac{1.5P}{P} = \frac{P * e^{0.5r}}{P}\) Simplifying, we get: \(1.5 = e^{0.5r}\) Step 2: Substitute the annual interest rate found in part (a): \(1.5 = e^{0.5 * 0.3466}\) Solving it, we have: \(1.5 = e^{0.1733}\) Step 3: Use the natural logarithm function (ln) to eliminate e from the equation: \(ln(1.5) = ln(e^{0.1733})\) Step 4: Use the property of logarithms that allows us to bring the exponent down: \(ln(1.5) = 0.1733 * ln(e)\) Since \(ln(e) = 1\), we have: \(ln(1.5) = 0.1733\) Step 5: We know that the original amount doubles when A = 2*P: \(2P = P * e^{rt}\) Dividing by P, we get: \(2 = e^{rt}\) Step 6: We can replace r with the value found in part (a): \(2 = e^{0.3466t}\) Step 7: Use the natural logarithm function (ln) to eliminate e from the equation: \(ln(2) = ln(e^{0.3466t})\) Step 8: Use the property of logarithms that allows us to bring the exponent down: \(ln(2) = 0.3466t * ln(e)\) Since \(ln(e) = 1\), we have: \(ln(2) = 0.3466t\) Step 9: Divide by 0.3466 to solve for t: \(t = \frac{ln(2)}{0.3466}\) The time t ≈ 2 years. It takes approximately 2 years for the original amount to double when it increases by 50% in six months.