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$Question 2 If R35 000 accumulates to R48 320 at a continuous compounding rate of $ 8,6 \% $ per year, determine the term under consideration in years to two d$

Question 2 If R35 000 accumulates to R48 320 at a continuous compounding rate of $ 8,6 \% $ per year, determine the term under consideration in years to two decimal places.

Ask by Moreno Collins.

Feb 16,2025 04:05

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Tutor-Verified Answer

Answer

The term under consideration is approximately 3.74 years.

Solution

For continuous compounding, the formula for the accumulated amount is: A = P × e^(r × t) Where: A = accumulated amount (R48,320) P = principal (R35,000) r = annual interest rate (0.086 per year) t = time in years Substitute the given values into the equation: 48,320 = 35,000 × e^(0.086t) 1. Divide both sides by 35,000: 48,320 / 35,000 = e^(0.086t) 2. Calculate the left-hand side: 1.38 (approximately) 3. Take the natural logarithm of both sides to solve for t: ln(1.38) = ln(e^(0.086t)) ln(1.38) = 0.086t 4. Solve for t: t = ln(1.38) / 0.086 5. Compute ln(1.38): ln(1.38) ≈ 0.322 6. Finally, divide by 0.086: t ≈ 0.322 / 0.086 ≈ 3.74 Thus, the term under consideration is approximately 3.74 years (to two decimal places).

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Bonus Knowledge

To find the term under consideration, we can use the formula for continuous compounding, which is given by: \[ A = Pe^{rt} \] where: - $ A $ is the final amount (R48,320), - $ P $ is the principal amount (R35,000), - $ r $ is the interest rate (8.6% or 0.086), and - $ t $ is the time in years. First, we rearrange the formula to solve for $ t $: \[ t = \frac{\ln(A/P)}{r} \] Plugging in the values: \[ t = \frac{\ln(48320 / 35000)}{0.086} \] Calculating further: \[ t = \frac{\ln(1.38)}{0.086} \approx \frac{0.324}{0.086} \approx 3.77 \] Thus, the term is approximately **3.77 years** when rounded to two decimal places.