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Chapter 3: Problem 15

Determine the time necessary for \(P\) dollars to double when it is invested at interest rate \(r\) compounded (a) annually, (b) monthly, (c) daily, and (d) continuously. $$r=10 \%$$

Short Answer

Expert verified
The time necessary for \(P\) dollars to double when it is invested at interest rate \(r\)=10\% compounded annually is approximately 7.27 years, monthly is approximately 6.98 years, daily is approximately 6.91 years, and continuously is approximately 6.93 years.

Step by step solution

01

Solving for annual compounding

For the case of annual compounding, interest is added once per year so \(n=1\). So the formula becomes: \(2P = P(1 + \frac{r}{1})^{1 \cdot t}\). Simplifying and solving for \(t\), we get \(t = \frac{\log(2)}{\log(1 + r)}\). With the given \(r = 10\%\) or \(0.1\), we find \(t \approx 7.27\) years.
02

Solving for monthly compounding

For the case of monthly compounding, interest is added 12 times per year so \(n=12\). We get: \(2P = P(1 + \frac{r}{12})^{12 \cdot t}\). After simplifying and solving for t, we get \(t \approx 6.98\) years.
03

Solving for daily compounding

For the case of daily compounding, interest is added 365 times per year so n=365. The formula will be: \(2P = P(1 + \frac{r}{365})^{365 \cdot t}\). After simplifying and solving for \(t\), we get \(t \approx 6.91\) years.
04

Solving for continuous compounding

For the case of continuous compounding, the interest is added an infinite number of times per year. The formula will be : \(2P = Pe^{rt}\). After simplifying, we solve for \(t\) to get \(t = \frac{\log(2)}{r} \approx 6.93\) years.

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