Question

how many years are required for an investment to double in value if it is appreciating at the rate of 5% compounded continuously?
at 5% compounded continuously, the investment doubles in
(round to one decimal place as needed.)

Explanation:

Step1: Use 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 number of years. If the investment doubles, then \(A = 2P\). Substituting into the formula gives \(2P=Pe^{rt}\). Since \(P
eq0\) (if \(P = 0\), there is no investment), we can divide both sides of the equation by \(P\). So, \(2=e^{rt}\). Given \(r = 0.05\) (because \(5\%=0.05\)), the equation becomes \(2 = e^{0.05t}\).

Step2: Take the natural logarithm of both sides

Take the natural logarithm of both sides of the equation \(2 = e^{0.05t}\). Using the property \(\ln(e^{x})=x\), we have \(\ln(2)=\ln(e^{0.05t})\). So, \(\ln(2)=0.05t\).

Step3: Solve for \(t\)

We know that \(\ln(2)\approx0.6931\). Then, \(t=\frac{\ln(2)}{0.05}\). Substitute \(\ln(2)\approx0.6931\) into the formula: \(t=\frac{0.6931}{0.05}=13.862\).

Answer:

\(13.9\)