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Chapter 4: Problem 65

The monthly payment \(P\) (in \(\$$ ) to pay off a loan of amount \)A\( (in \)\$$ ) at an interest rate \(r\) in \(t\) years is given by $$P=\frac{\frac{A r}{12}}{1-\left(1+\frac{r}{12}\right)^{-12 t}}$$ a. Solve for \(t\) (note that there are numerous equivalent algebraic forms for the result). b. Interpret the meaning of the resulting relationship.

Short Answer

Expert verified
To find the time \(t\): \[t = \frac{\ln \left(\frac{P - \frac{A r}{12}}{P}\right)}{-12 \ln \left(1+\frac{r}{12}\right)}\]. This shows how monthly payments, loan amount, and interest rate affect the duration of the loan.

Step by step solution

01

Identify the formula and given values

The given formula for the monthly payment is: \[P = \frac{\frac{A r}{12}}{1-\left(1+\frac{r}{12}\right)^{-12 t}}\] We are tasked with solving for \(t\).
02

Isolate the fraction involving time

Multiply both sides of the equation by the denominator to isolate the fraction involving time: \[P \left(1-\left(1+\frac{r}{12}\right)^{-12 t}\right) = \frac{A r}{12}\]
03

Distribute and isolate the exponential term

Distribute the term \(P\), then isolate the exponential term: \[P - P \left(1+\frac{r}{12}\right)^{-12 t} = \frac{A r}{12}\]\[-P \left(1+\frac{r}{12}\right)^{-12 t} = \frac{A r}{12} - P\]
04

Solve for the exponential term

Isolate the exponential term by dividing both sides by \(-P\): \[\left(1+\frac{r}{12}\right)^{-12 t} = \frac{P - \frac{A r}{12}}{P}\]
05

Take the natural logarithm (ln) of both sides

To solve for \(t\), take the natural logarithm of both sides:\[\ln \left(\left(1+\frac{r}{12}\right)^{-12 t}\right) = \ln \left(\frac{P - \frac{A r}{12}}{P}\right)\]
06

Simplify using logarithm properties

Apply the property of logarithms, \(\ln(a^b) = b \ln(a)\), to simplify: \[-12 t \ln \left(1+\frac{r}{12}\right) = \ln \left(\frac{P - \frac{A r}{12}}{P}\right)\]
07

Solve for time (t)

Isolate \(t\) by dividing both sides by \(-12 \ln \left(1+\frac{r}{12}\right)\): \[t = \frac{\ln \left(\frac{P - \frac{A r}{12}}{P}\right)}{-12 \ln \left(1+\frac{r}{12}\right)}\]
08

Final relationship interpretation

The relationship shows that the time \(t\) to fully pay off a loan depends on the monthly payment \(P\), loan amount \(A\), and interest rate \(r\). Specifically, if the monthly payment is higher, or the interest rate is lower, the time required to pay off the loan decreases.

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Key Concepts

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

loan repayment formula
When you borrow money, like a student loan or a mortgage, you need to pay it back with some extra amount called interest. One common way to calculate the monthly payment is through a formula:

The monthly payment, denoted as \(P\), is given by: \[P = \frac{\frac{A r}{12}}{1-\left(1+\frac{r}{12}\right)^{-12t}}\] Here:
  • \(A\) is the total amount borrowed (loan principal).
  • \(r\) is the annual interest rate (expressed as a decimal).
  • \(t\) is the time in years to fully pay off the loan.
This formula helps you figure out how much you need to pay every month to clear your loan within a certain period. Breaking down the formula step-by-step can make understanding how it works easier.
solving exponential equations
Exponential equations involve terms where the variable is in an exponent. In our formula, the tricky part is the term \((1 + \frac{r}{12})^{-12t}\).

Here鈥檚 how to solve these types of equations:
  • First, isolate the exponential term.
  • Then, take the natural logarithm (ln) of both sides to remove the exponent.
  • Finally, solve for the variable.
For instance, in our equation, after isolating the exponential term, we apply: \[ \ln \left( (1 + \frac{r}{12})^{-12t} \right)\] Using the logarithm property \( \ln (a^b) = b \ln (a) \), you can simplify it to: \[-12t \ln (1 + \frac{r}{12})\]This simplification helps in solving for the time \(t\).
logarithms in algebra
Logarithms are the inverses of exponentials. They play a crucial role in simplifying exponential equations, especially in loan repayment calculations.

For example, if you have: \[10^2 = 100\] Taking the logarithm (base 10) of both sides gives: \[\log_{10}(10^2) = \log_{10}(100)\] Which simplifies to: \[2 = \log_{10}(100)\] In loan calculations, the natural logarithm (ln) is used because it's based on the constant \(e\). For an exponential term \(a = e^b\), the natural logarithm simplifies as: \[\ln(a) = b\] So, taking the natural logarithm of both sides of our isolated exponential term: \[\ln \left( (1 + \frac{r}{12})^{-12t} \right)\] Helps to solve for time \(t\).
interest rate impact
The interest rate \(r\) is a crucial factor in loan repayment. It determines how much extra you pay on top of the amount borrowed. Its impact on the loan payment can be observed in several ways:

  • Higher interest rates result in higher monthly payments.
  • Lower interest rates make repaying the loan faster, as more of each payment goes towards the principal instead of interest.
  • Even a small change in interest rate can significantly affect the repayment time and total amount paid.
In our formula, \(r\) is involved in both the numerator and the denominator: \[P = \frac{\frac{A r}{12}}{1-\left(1+\frac{r}{12}\right)^{-12t}}\] A lower \(r\) decreases the \( \frac{A r}{12}\) term, making the monthly payment lower. Understanding the impact of the interest rate helps in making smarter borrowing decisions.

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Most popular questions from this chapter

9\. The population of the United States \(P(t)\) (in millions) since January 1,1900 , can be approximated by $$P(t)=\frac{725}{1+8.295 e^{-0.0165 t}}$$ where \(t\) is the number of years since January \(1,1900 .\) (See Example 6\()\) a. Evaluate \(P(0)\) and interpret its meaning in the context of this problem. b. Use the function to predict the U.S. population on January \(1,2020 .\) Round to the nearest million. c. Use the function to predict the U.S. population on January 1,2050 . d. Determine the year during which the U.S. population will reach 500 million. e. What value will the term \(\frac{8.295}{e^{0.0165 t}}\) approach as \(t \rightarrow \infty\) ? f. Determine the limiting value of \(P(t)\).

A function of the form \(P(t)=a b^{t}\) represents the population of the given country \(t\) years after January 1,2000 . a. Write an equivalent function using base \(e\); that is, write a function of the form \(P(t)=P_{0} e^{k t} .\) Also, determine the population of each country for the year 2000 . $$\begin{array}{|l|c|c|c|} \hline \text { Country } & P(t)=a b^{t} & P(t)=P_{0} e^{k t} & \begin{array}{c} \text { Population } \\ \text { in } 2000 \end{array} \\ \hline \text { Haiti } & P(t)=8.5(1.0158)^{t} & & \\ \hline \text { Sweden } & P(t)=9.0(1.0048)^{t} & & \\ \hline \end{array}$$ b. The population of the two given countries is very close for the year 2000 , but their growth rates are different. Determine the year during which the population of each country will reach 10.5 million. c. Haiti had fewer people in the year 2000 than Sweden. Why did Haiti reach a population of 10.5 million sooner?

The population of Canada \(P(t)\) (in millions) since January \(1,1900,\) can be approximated by $$P(t)=\frac{55.1}{1+9.6 e^{-0.02515 t}}$$ where \(t\) is the number of years since January 1,1900 . a. Evaluate \(P(0)\) and interpret its meaning in the context of this problem. b. Use the function to predict the Canadian population on January \(1,2015 .\) Round to the nearest million. c. Use the function to predict the Canadian population on January 1,2040 . d. Determine the year during which the Canadian population will reach 45 million. e. What value will the term \(\frac{9.6}{e^{0.02515 t}}\) approach as \(t \rightarrow \infty\) ? f. Determine the limiting value of \(P(t)\).

A table of data is given. a. Graph the points and from visual inspection, select the model that would best fit the data. Choose from $$\begin{array}{ll} y=m x+b \text { (linear) } & y=a b^{x} \text { (exponential) } \\ y=a+b \ln x \text { (logarithmic) } & y=\frac{c}{1+a e^{-b x}} \text { (logistic) } \end{array}$$ b. Use a graphing utility to find a function that fits the data. $$ \begin{array}{|c|c|} \hline x & y \\ \hline 3 & 2.7 \\ \hline 7 & 12.2 \\ \hline 13 & 25.7 \\ \hline 15 & 30 \\ \hline 17 & 34 \\ \hline 21 & 44.4 \\ \hline \end{array} $$

Which functions are exponential? a. \(f(x)=\left(\frac{1}{\sqrt{3}}\right)^{x}\) b. \(f(x)=1^{x}\) c. \(f(x)=x^{\sqrt{3}}\) d. \(f(x)=(-2)^{x}\) e. \(f(x)=\pi^{x}\)
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