Question

Find the time necessary for \(\$ 1000\) to double if it is invested at a rate of \(r\) compounded (a) annually, (b) monthly, (c) daily, and (d) continuously. $$ r=7 \% $$

Short answer

The time necessary for \$1000 to double if it is invested at a rate of 7% per annum is approximately 10.24 years when compounded annually, 9.99 years when compounded monthly, 9.96 years when compounded daily and 9.92 years when compounded continuously.

Step-by-step solution

Solving for t when compounded annually

The formula \(A = P(1 + r/n)^{nt}\) becomes \(2000 = 1000(1 + 0.07/1)^{1*t}\). After simplifying we get \(2 = (1.07)^t\). Taking the natural logarithm of both sides gives \(\ln(2) = t \ln(1.07)\) which simplifies to \(t = \frac{\ln(2)}{\ln(1.07)} \approx 10.24\). Thus it takes approximately 10.24 years for the money to double.

Solving for t when compounded monthly

The formula becomes \(2000 = 1000(1 + 0.07/12)^{12*t}\). Following the same process as before gives \(t = \frac{\ln(2)}{12 \ln(1 + 0.07/12)} \approx 9.99\). It takes approximately 9.99 years for the money to double.

Solving for t when compounded daily

The formula becomes \(2000 = 1000(1 + 0.07/365)^{365*t}\). Following the same process as before gives \(t = \frac{\ln(2)}{365 \ln(1 + 0.07/365)} \approx 9.96\). It takes approximately 9.96 years for the money to double.

Solving for t when compounded continuously

We use the continuous compounding formula \(A = Pe^{rt}\) which becomes \(2000 = 1000e^{0.07t}\). Taking the natural logarithm of both sides gives \(\ln(2) = 0.07t\) which simplifies to \(t = \frac{\ln(2)}{0.07} \approx 9.92\). It takes approximately 9.92 years for the money to double.