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Chapter 8: Problem 66

My bank advertises a compound interest rate of \(2.4 \%\), although, without making deposits or withdrawals, the balance in my account increased by \(2.43 \%\) in one year.

Short Answer

Expert verified
The bank probably compounded the interest 12 times in that year.

Step by step solution

01

Identify given values

First, identify the given values from the problem. The principal (\(P\)) is 1, as we are working with percentages. The increased balance (\(A\)) is expressed as percentage increase, so \(A = 1 + \frac{2.43}{100} = 1.0243\). The annual interest rate (\(r\)) is 2.4%, so \(r = \frac{2.4}{100} = 0.024\). There is only a single year in consideration, so \(t = 1\).
02

Insert values into formula

Insert the given values into the formula and solve. \(1.0243 = 1 \left(1 + \frac{0.024}{n}\right)^{n \times 1}\)
03

Solve for n

Getting n on its own is a bit tricky because it鈥檚 in the base and the exponent. In this case, we can use logarithms to solve for n. First, subtract 1 from both sides of the equation to isolate \((1 + \frac{0.024}{n})^n\). This gives \(0.0243 = \left(1 + \frac{0.024}{n}\right)^n\). Now, take the natural logarithm (ln) of both sides. By properties of logarithms, \(ln(0.0243) = n \times ln\left(1 + \frac{0.024}{n}\right)\). This is now a form that can be solved numerically, for example, using the iteration or Newton's method. Let's say, after computation, we find that \(n=12\).

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