Question

$600 were deposited into an account with a 5% interest rate, compounded continuously. how many years was it in the bank if the current amount is $6000? t = ? years

Explanation:

Step1: Write the continuous - compounding formula

The formula for continuous compounding is \(A = Pe^{rt}\), where \(A\) is the final amount, \(P\) is the principal amount, \(r\) is the annual interest rate (in decimal form), and \(t\) is the time in years.
Given \(P=\$600\), \(A = \$6000\), and \(r=0.05\). Substitute these values into the formula: \(6000=600e^{0.05t}\).

Step2: Simplify the equation

Divide both sides of the equation by \(600\):
\(\frac{6000}{600}=e^{0.05t}\), which simplifies to \(10 = e^{0.05t}\).

Step3: Take the natural logarithm of both sides

Using the property \(\ln(e^{x})=x\), take the natural logarithm of both sides: \(\ln(10)=\ln(e^{0.05t})\).
Since \(\ln(e^{0.05t}) = 0.05t\), the equation becomes \(\ln(10)=0.05t\).

Step4: Solve for \(t\)

We know that \(\ln(10)\approx2.3026\). Then \(t=\frac{\ln(10)}{0.05}\).
Substitute \(\ln(10)\approx2.3026\) into the formula: \(t=\frac{2.3026}{0.05}=46.052\).

Answer:

\(t = 46.05\) years (rounded to two decimal places)