Question

complete the table for a savings account subject to continuous compounding (a = pe^{rt}).

amount invested	annual interest rate	accumulated amount	time t in years

(round to the nearest whole number as needed).

amount invested	annual interest rate	accumulated amount	time t in years

Explanation:

Step1: Recall the continuous compounding formula

The formula for continuous compounding is \( A = Pe^{rt} \), where \( A \) is the accumulated amount, \( P \) is the principal (amount invested), \( r \) is the annual interest rate (in decimal), and \( t \) is the time in years.

We know that \( P = 2100 \), \( r = 22.0\% = 0.22 \), and \( A = 3P \) (since we want to triple the amount invested, so \( A = 3 \times 2100 = 6300 \)).

Substitute these values into the formula: \( 6300 = 2100e^{0.22t} \)

Step2: Solve for \( t \)

First, divide both sides by 2100:
\( \frac{6300}{2100} = e^{0.22t} \)
\( 3 = e^{0.22t} \)

Take the natural logarithm of both sides:
\( \ln(3) = \ln(e^{0.22t}) \)

Using the property \( \ln(e^x) = x \), we get:
\( \ln(3) = 0.22t \)

Now, solve for \( t \) by dividing both sides by 0.22:
\( t = \frac{\ln(3)}{0.22} \)

Calculate \( \ln(3) \approx 1.0986 \), so:
\( t \approx \frac{1.0986}{0.22} \approx 4.9936 \)

Round to the nearest whole number, \( t \approx 5 \)

Answer:

5