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Chapter 4: Problem 18

About how many years does it take for money to double when compounded continuously at \(10 \%\) per year?

Short Answer

Expert verified
It takes approximately \(6.93\) years for the money to double when compounded continuously at a \(10\%\) annual interest rate.

Step by step solution

01

Write down the formula for continuous compounding

First, write the formula for continuous compounding: \(A = Pe^{rt}\), where A is the final amount, P is the principal amount, r is the interest rate, and t is the time in years.
02

Set the final amount A to 2P

Since the money is doubling, the final amount A equals 2 times the principal amount P: \(2P = Pe^{rt}\).
03

Divide both sides of the equation by P

Divide both sides of the equation by P to isolate the exponential term: \(\frac{2P}{P} = \frac{Pe^{rt}}{P}\). This simplifies to \(2 = e^{rt}\).
04

Take the natural logarithm of both sides of the equation

To solve for t, take the natural logarithm of both sides of the equation: \(\ln{2} = \ln{(e^{rt})}\).
05

Use the power rule of logarithms

The power rule of logarithms states that \(\ln{(e^{rt})} = rt\). Therefore, \(\ln{2} = rt\).
06

Substitute the interest rate r with its decimal equivalent

The interest rate is given as 10% per year. In decimal form, r = 0.10.
07

Solve for t

Now, replace r with its decimal value in the equation: \(\ln{2} = 0.10t\). Divide both sides of the equation by 0.10 to solve for t: \(t = \frac{\ln{2}}{0.10}\).
08

Calculate the time it takes for the money to double

Using a calculator, compute the value of \(\frac{\ln{2}}{0.10}\): \(t \approx 6.93\). It takes approximately 6.93 years for the money to double when compounded continuously at a 10% annual interest rate.

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