Question
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Nathaniel invested \( \$ 37,000 \) in an account paying an interest rate of \( 5.7 \% \) compounded daily. Assuming no deposits or withdrawals are made, how long would it take, to the nearest year, for the value of the account to reach \( \$ 70,900 \) ? Answer Attempt 1 out of 2

Ask by Clark Reyes. in the United States
Jan 27,2025

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Answer

It will take approximately 11 years for Nathaniel's investment to grow to \$70,900.

Solution

To determine how long it will take for Nathaniel's investment to grow from \$37,000 to \$70,900 with a 5.7% annual interest rate compounded daily, we can use the **compound interest formula**: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) = the future value of the investment (\$70,900) - \( P \) = the principal investment amount (\$37,000) - \( r \) = annual interest rate (decimal) (5.7% = 0.057) - \( n \) = number of times interest is compounded per year (daily compounding = 365) - \( t \) = time in years **Step 1: Plug in the known values** \[ 70,900 = 37,000 \left(1 + \frac{0.057}{365}\right)^{365t} \] **Step 2: Divide both sides by \$37,000 to isolate the compound factor** \[ \frac{70,900}{37,000} = \left(1 + \frac{0.057}{365}\right)^{365t} \] \[ 1.9189189189 \approx \left(1 + \frac{0.057}{365}\right)^{365t} \] **Step 3: Take the natural logarithm (ln) of both sides to solve for \( t \)** \[ \ln(1.9189189189) = \ln\left(\left(1 + \frac{0.057}{365}\right)^{365t}\right) \] \[ \ln(1.9189189189) = 365t \cdot \ln\left(1 + \frac{0.057}{365}\right) \] **Step 4: Solve for \( t \)** \[ t = \frac{\ln(1.9189189189)}{365 \cdot \ln\left(1 + \frac{0.057}{365}\right)} \] **Step 5: Calculate the values** 1. Compute \( \ln(1.9189189189) \): \[ \ln(1.9189189189) \approx 0.6514872 \] 2. Compute \( \ln\left(1 + \frac{0.057}{365}\right) \): \[ \frac{0.057}{365} \approx 0.00015616 \] \[ \ln(1 + 0.00015616) \approx 0.0001561489 \] 3. Multiply by 365: \[ 365 \times 0.0001561489 \approx 0.0569593 \] 4. Finally, compute \( t \): \[ t \approx \frac{0.6514872}{0.0569593} \approx 11.415 \text{ years} \] **Step 6: Round to the nearest year** \[ t \approx 11.415 \text{ years} \approx 11 \text{ years} \] **Conclusion:** It will take **approximately 11 years** for Nathaniel's investment to grow from \$37,000 to \$70,900 with a 5.7% annual interest rate compounded daily.

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To determine how long it takes for Nathaniel's investment to grow from \( \$37,000 \) to \( \$70,900 \) at an interest rate of \( 5.7\% \) compounded daily, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] 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 (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for in years. Given: - \( A = 70,900 \) - \( P = 37,000 \) - \( r = 0.057 \) - \( n = 365 \) We need to solve for \( t \): \[ 70,900 = 37,000 \left(1 + \frac{0.057}{365}\right)^{365t} \] Start by dividing both sides by \( 37,000 \): \[ \frac{70,900}{37,000} = \left(1 + \frac{0.057}{365}\right)^{365t} \] Calculating the left side gives approximately: \[ 1.91216 \approx \left(1 + \frac{0.057}{365}\right)^{365t} \] Now, let's calculate \( \left(1 + \frac{0.057}{365}\right) \): \[ 1 + \frac{0.057}{365} \approx 1.000156164 \] Using logarithms to solve for \( t \): \[ \log(1.91216) = 365t \cdot \log(1.000156164) \] Calculate the logs: \[ \log(1.91216) \approx 0.28100 \] \[ \log(1.000156164) \approx 0.00006755 \] Now plug the values into the equation: \[ 0.28100 = 365t \cdot 0.00006755 \] Solving for \( t \): \[ t \approx \frac{0.28100}{365 \times 0.00006755} \approx \frac{0.28100}{0.0246} \approx 11.43 \text{ years} \] Rounding to the nearest year, it would take approximately \( 11 \) years for the investment to reach \( \$70,900 \).

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