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Chapter 4: Problem 6

A savings account: You initially invest \(\$ 500\) in a savings account that pays a yearly interest rate of \(4 \%\). a. Write a formula for an exponential function giving the balance in your account as a function of the time since your initial investment. b. What monthly interest rate best represents this account? Round your answer to three decimal places. c. Calculate the decade growth factor. d. Use the formula you found in part a to determine how long it will take for the account to reach \(\$ 740\). Explain how this is consistent with your answer to part \(\mathrm{c}\).

Short Answer

Expert verified
The balance formula is \( B(t) = 500(1.04)^t \). The monthly interest rate is approximately 0.3273%. It takes around 10 years to reach $740.

Step by step solution

01

Determine the Exponential Function Formula

The formula for the balance in a savings account that earns interest compounded annually is given by the function \[ B(t) = P(1 + r)^t \] where \( B(t) \) is the balance after \( t \) years, \( P \) is the initial principal amount, and \( r \) is the annual interest rate. For this problem, \( P = \$500 \) and \( r = 0.04 \), thus the function becomes \[ B(t) = 500(1 + 0.04)^t = 500(1.04)^t \].
02

Find the Monthly Interest Rate

To convert the annual interest rate to a monthly interest rate, we use the formula for equivalent monthly compounding: \[ (1 + r)^{1/12} - 1 \]. Substituting the value of \( r = 0.04 \), we have \[ (1 + 0.04)^{1/12} - 1 = 1.04^{1/12} - 1 \]. Calculating this gives a monthly interest rate of approximately \( 0.003273 \) or \( 0.3273\% \).
03

Calculate the Decade Growth Factor

The decade growth factor is the amount the original investment grows after 10 years. Using the formula from Step 1, we find \[ B(10) = 500(1.04)^{10} \]. Calculating \( (1.04)^{10} \) gives approximately \( 1.48024 \). Thus, the decade growth factor is \( 1.48024 \), meaning the balance grows by a factor of 1.48024 over 10 years.
04

Solve for Time to Reach $740

To determine how long it will take for the balance to reach \$740, we set \[ 740 = 500(1.04)^t \] and solve for \( t \). Dividing both sides by 500 gives \[ (1.04)^t = \frac{740}{500} = 1.48 \]. Taking the natural logarithm of both sides, \[ \ln((1.04)^t) = \ln(1.48) \]. Using the properties of logarithms, this becomes \[ t \cdot \ln(1.04) = \ln(1.48) \]. Solving for \( t \) gives \[ t = \frac{\ln(1.48)}{\ln(1.04)} \approx 10 \] years. This matches the decade growth factor calculated in Step 3, showing consistency as the account indeed grows by a factor of \( 1.48 \) over 10 years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interest Rates
Interest rates are a key concept in understanding financial growth. Whenever you deposit money in a bank account or take out a loan, an interest rate will either be building your savings or adding to your debt. An interest rate is typically expressed as an annual percentage. For example, if you have a savings account with a 4% annual interest rate, this means that for each year, your savings can grow by 4% of the initial amount.
  • Interest rates can vary from bank to bank and across different financial products.
  • They have direct impacts on how quickly savings can grow or how loan balances increase over time.
Understanding the interest rate gives you a clearer picture of how your money will develop over time.
Compounded Interest
Compounded interest is what makes your money in a savings account grow faster compared to simple interest. Unlike simple interest where interest is calculated only on the initial principal, compounded interest adds interest to the principal, then calculates interest on this new amount.
For example, if your account compounds annually at 4%, you earn interest not just on your initial $500, but also on the interest previously earned.
  • Compounded interest can occur on different time scales, such as annually, semi-annually, quarterly, monthly, etc.
  • The frequency of compounding will affect how much total interest you earn.
The more frequently interest is compounded, the more your money grows.
Financial Modeling
Financial modeling is the process of creating a mathematical representation of a financial situation. This is done to predict future financial performance based on historical data. In personal finance, simple exponential functions can be used as models to predict savings growth over time.
The formula used in the exercise, \( B(t) = 500(1.04)^t \), is a straightforward model showing how an investment grows with compounded interest.
  • Financial models can help in making informed decisions about savings, investments, and loans.
  • They serve as a planning tool to gauge future financial scenarios.
Using such models, you can develop strategies to meet long-term financial goals.
Decade Growth Factor
The decade growth factor shows how much an initial investment grows over a ten-year period. It is an easy way to see the long-term power of compounded interest.
In the exercise, the decade growth factor is calculated using the formula \( B(10) = 500(1.04)^{10} \), where \( (1.04)^{10} \) approximates to 1.48024.
  • The decade growth factor here is about 1.48, meaning that the money grows by nearly 48%.
  • This calculation provides a quick insight into how effective an investment is over a significant period.
Understanding the decade growth factor helps in visualizing how your finances can evolve in the long run.

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Most popular questions from this chapter

Magazine sales: The following table shows the income from sales of a certain magazine, measured in thousands of dollars, at the start of the given year. $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Income } \\ \hline 2001 & 7.76 \\ \hline 2002 & 8.82 \\ \hline 2003 & 9.88 \\ \hline 2004 & 10.94 \\ \hline 2005 & 12.00 \\ \hline 2006 & 13.08 \\ \hline 2007 & 14.26 \\ \hline 2008 & 15.54 \\ \hline \end{array} $$ Over an initial period the sales grew at a constant rate, and over the rest of the time the sales grew at a constant percentage rate. Calculate differences and ratios to determine what these time periods are, and find the growth rate or percentage growth rate, as appropriate.

Exponential decay with given initial value and decay factor: Write the formula for an exponential function with initial value 200 and decay factor 0.73. Plot its graph.

Headway on four-lane highways: When traffic is flowing on a highway, the headway is the average time between vehicles. On four-lane highways, the probability \(P\) that the headway is at least \(t\) seconds is given to a good degree of accuracy \({ }^{6}\) by $$ P=e^{-q t}, $$ a. On a four-lane highway carrying an average of 500 vehicles per hour in one direction, what is the probability that the headway is at least 15 seconds? (Note: 500 vehicles per hour is \(\frac{500}{3600}=0.14\) vehicle per second.) b. On a four-lane highway carrying an average of 500 vehicles per hour, what is the decay factor for the probability that headways are at least \(t\) seconds? Reminder: An important law of exponents tells us that \(a^{b c}=\left(a^{b}\right)^{c}\). where \(q\) is the average number of vehicles per second traveling one way on the highway.

Cell phones: The following table shows the number, in millions, of cell phone subscribers in the United States at the end of the given year. $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Subscribers (millions) } \\ \hline 2001 & 128.4 \\ \hline 2002 & 140.8 \\ \hline 2003 & 158.7 \\ \hline 2004 & 182.1 \\ \hline 2005 & 207.9 \\ \hline \end{array} $$ a. Plot the data points. Does this plot make it look reasonable to approximate the data with an exponential function? b. Use exponential regression to construct an exponential model for the subscriber data. c. Add the graph of the exponential model to the plot in part a. d. What was the yearly percentage growth rate from the end of 2001 through the end of 2005 for cell phone subscribership? e. In 2005 an executive had a plan that could make money for the company, provided that there would be at least 250 million cell phone subscribers by the end of 2007. Solely on the basis of an exponential model for the data in the table, would it be reasonable for the executive to implement the plan?

Sound pressure: Sound exerts pressure on the human ear. Increasing loudness corresponds to greater pressure. The table below shows the pressure P , in dynes per square centimeter, exerted on the ear by sound with loudness D, measured in decibels. $$ \begin{array}{|c|c|} \hline \text { Loudness } D & \text { Pressure } P \\ \hline 65 & 0.36 \\ \hline 85 & 3.6 \\ \hline 90 & 6.4 \\ \hline 105 & 30 \\ \hline 110 & 50 \\ \hline \end{array} $$ a. Plot the natural logarithm of the data. Does it appear reasonable to model pressure as an exponential function of loudness? b. Find an exponential model of P as a function of D using the logarithm as a link. c. How is pressure on the ear affected when loudness is increased by 1 decibel?
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