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Chapter 4: Problem 71

If monthly payments \(p\) are deposited in a savings account paying an annual interest rate \(r,\) then the amount \(A\) in the account after \(n\) years is given by $$A=\frac{p\left(1+\frac{r}{12}\right)\left[\left(1+\frac{r}{12}\right)^{12 n}-1\right]}{\frac{r}{12}}$$ Graph \(A\) for each value of \(p\) and \(r,\) and estimate \(n\) for \(A=100,000 \text{dollars}\). $$p=100, \quad r=0.05$$

Short Answer

Expert verified
The account reaches $100,000 in approximately 38.45 years with $100 monthly deposits and a 5% annual interest rate.

Step by step solution

01

Understand the formula

The formula given is a mathematical expression for estimating the total amount A in a savings account over \(n\) years, with monthly deposits \(p\), at an annual interest rate \(r\). It accounts for compound interest applied monthly. The critical variables are \(p\), \(r\), and \(n\).
02

Plug known values into the formula

We are given \(p = 100\) and \(r = 0.05\). Plug these values into the formula:\[A = \frac{100\left(1 + \frac{0.05}{12}\right)\left[\left(1 + \frac{0.05}{12}\right)^{12n} - 1\right]}{\frac{0.05}{12}}\]
03

Simplify constants

Calculate constants within the formula first. Compute \(\frac{0.05}{12} = 0.0041667\) and the monthly interest multiplier \(1 + 0.0041667 = 1.0041667\). Substitute these back:\[A = \frac{100(1.0041667)\left[\left(1.0041667\right)^{12n} - 1\right]}{0.0041667}\]
04

Solve for A when it equals 100,000 dollars

Set \(A = 100,000\) and solve for \(n\):\[100,000 = \frac{100 \cdot 1.0041667 \cdot \left((1.0041667)^{12n} - 1\right)}{0.0041667}\]Re-arrange to find \(n\):\[1.0041667 \left((1.0041667)^{12n} - 1\right) = 416.67\]\[\left((1.0041667)^{12n} - 1\right) = 416.67 / 1.0041667\approx 414.17\]\[(1.0041667)^{12n} \approx 415.17\]
05

Use logarithms to solve for n

Take the natural logarithm of both sides to solve for \(n\):\[\ln((1.0041667)^{12n}) = \ln(415.17)\]Use the logarithm power rule:\[12n \cdot \ln(1.0041667) = \ln(415.17)\]\[n = \frac{\ln(415.17)}{12 \cdot \ln(1.0041667)}\]Solve numerically to get:\[n \approx 38.45\]
06

Interpret the result

Since \(n\) is the number of years needed to reach \(A = 100,000\) with \(p = 100\) monthly deposits at an interest rate of \(r = 0.05\), you will need about \(38.45\) years. This emphasizes the long-term nature of saving with regular small deposits and compound interest.

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Key Concepts

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

Savings Growth Calculation
Understanding how savings grow over time is essential for effective financial planning. The formula we are working with calculates the future value of a savings account through monthly deposits, considering compound interest. Compound interest means the interest applied not only to the initial principal but also to the accumulated interest from previous periods. This effect means your savings grow faster over time.To calculate this, we use the following formula:\[A = \frac{p\left(1+\frac{r}{12}\right)\left[\left(1+\frac{r}{12}\right)^{12n}-1\right]}{\frac{r}{12}}\]Where:- \(A\) is the total amount in the account after \(n\) years.- \(p\) is the monthly deposit amount.- \(r\) stands for the annual interest rate.- Monthly compounding is accounted for by \(\frac{r}{12}\).By plugging the relevant values such as your monthly deposit \(p\) and annual interest \(r\) into the equation, you can forecast how much your savings will grow over a specified period \(n\). This empowers you to set realistic saving goals and adjust as necessary.
Time Value of Money
The Time Value of Money (TVM) principle explains that a specific amount of money today is worth more than the same amount in the future, due to its potential earning capacity. This concept is fundamentally tied to savings and investments.By applying this principle, it's clear that starting to save earlier can greatly increase the amount accumulated over time, as each dollar has more time to earn interest. In the savings growth calculation, regular monthly deposits increase the opportunity for compound interest, reflecting TVM.In our example, saving \(p = 100\) monthly, into an account with a 5% annual interest rate allows the initial deposits to earn interest not only on the principal but also on the accumulated interest. This highlights the critical role of time in compounding:
  • The earlier you start saving, the more time your money has to grow.
  • The effect of compound interest grows over time, significantly boosting your savings.
Understanding TVM is vital for making informed financial decisions, such as comparing investment opportunities or assessing future financial needs.
Logarithmic Functions
Logarithmic functions are invaluable in solving equations involving exponential growth, particularly when you need to find the time required to reach a specific future value of savings. In our exercise, logs help determine the number of years \(n\) required for the savings account to grow to \(A = 100,000\) with given monthly contributions and interest rate.Here's how logarithms come into play:Often, savings growth equations result in expressions where the term inside the exponential part, like \((1.0041667)^{12n}\), must be isolated. To bring down an exponent and solve for \(n\), take the natural logarithm of both sides. This is critical because:
  • The logarithm helps translate compounded growth into a linear equation.
  • We can solve for \(n\) using algebraic manipulation of the equation once simplified with logs.
  • It allows you to understand how small monthly increments and percentage changes compound over decades.
This method offers a precise approach to unravel complex calculations in finance, enabling you to pinpoint the exact time needed to hit targeted savings goals.
Monthly Compounding Interest
Monthly compounding interest significantly enhances the growth of savings compared to annually compounded interest. It involves calculating interest every month and adding it to the account balance, meaning you earn interest on new deposits and on accrued interest. This effect results in faster capital growth.In our formula, the component \(\left(1+\frac{r}{12}\right)\) reflects monthly compounding:
  • Interest is calculated each month and is represented by \(\frac{r}{12}\), where \(r\) is the annual rate.
  • By dividing the annual rate by 12 and compounding monthly, the formula adjusts the growth pattern to reflect more frequent interest credits.
  • This can lead to a substantial increase in the final account balance over many years, compared to less frequent compounding.
The benefits of monthly compounding illustrate how even small changes in the frequency of interest application can significantly impact long-term savings. Therefore, when evaluating savings accounts, considering whether interest is compounded monthly or otherwise can reveal much about potential future growth.

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Most popular questions from this chapter

Cooling A jar of boiling water at \(212^{\circ} \mathrm{F}\) is set on a table in a room with a temperature of \(72^{\circ} \mathrm{F}\). If \(T(t)\) represents the temperature of the water after \(t\) hours, graph \(T(t)\) and determine which function best models the situation. (1) \(T(t)=212-50 t\) (2) \(T(t)=140 e^{-t}+72\) (3) \(T(t)=212 e^{-t}\) (4) \(T(t)=72+10 \ln (140 t+1)\)

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