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Chapter 4: Problem 5

Suppose a bank account that compounds interest continuously grows from \(\$ 100\) to \(\$ 110\) in two years. What annual interest rate is the bank paying?

Short Answer

Expert verified
The annual interest rate the bank is paying is approximately 4.879%. We found this by using the continuous compounding interest formula and solving for the interest rate \(r\).

Step by step solution

01

Substitute the Given Values in the Formula

Replace A, P, and t with the given values in the continuous compounding interest formula: \[110 = 100 * e^{2r}\]
02

Isolate the Exponential Term

Divide both sides of the equation by 100 to isolate the exponential term on the right side of the equation: \[\frac{110}{100} = e^{2r}\]
03

Simplify the Equation

Simplify the left side of the equation: \[1.1 = e^{2r}\]
04

Take the Natural Logarithm of Both Sides

Take the natural logarithm (ln) of both sides of the equation to get rid of the exponential term and solve for r: \[ln(1.1) = ln(e^{2r})\]
05

Use the Property of Natural Logarithms

Using the property ln(a^b) = b * ln(a), we can write: \[ln(1.1) = 2r * ln(e)\] Since ln(e) = 1, we can simplify our equation to: \[ln(1.1) = 2r\]
06

Solve for r

Divide both sides of the equation by 2 to solve for the interest rate r: \[r = \frac{ln(1.1)}{2}\]
07

Calculate the Value of r

Use a calculator to find the value of r (as a decimal): \[r \approx 0.04879\]
08

Convert r to Percentage

To express the interest rate r as a percentage, multiply the decimal value by 100: \[Annual\, Interest\, Rate \approx 0.04879 * 100\] \[Annual\, Interest\, Rate \approx 4.879\%\] The annual interest rate that the bank is paying is approximately 4.879%.

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