/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 You invest \(\$ 2500\) in an acc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 72

You invest \(\$ 2500\) in an account at interest rate \(r,\) compounded continuously. Find the time required for the amount to (a) double and (b) triple. $$r=0.0375$$

Short Answer

Expert verified
The time required for the investment to double is approximately \(\frac{ln(2)}{0.0375}\) years and to triple is approximately \(\frac{ln(3)}{0.0375}\) years.

Step by step solution

01

Understand the continuous compound interest formula

The continuous compound interest formula is \(A = P e^{rt}\). Here, \(A\) represents the final amount, \(P\) represents the initial principle, \(r\) represents the annual interest rate, and \(t\) represents the time in years.
02

Solve for time (t) to double the initial principle

We substitute the given values into the formula and solve for \(t\). In this case, \(P= 2500\), \(A = 2P = 5000\), and \(r = 0.0375\). The equation becomes \(5000 = 2500 e^{0.0375t}\). Solving this gives \(t = \frac{ln(\frac{5000}{2500})}{0.0375} = \frac{ln(2)}{0.0375}.\)
03

Solve for time (t) to triple the initial principle

Similarly, we substitute the given values into the formula and solve for \(t\). In this case, \(P= 2500\), \(A = 3P = 7500\), and \(r = 0.0375\). The equation becomes \(7500 = 2500 e^{0.0375t}\). Solving this gives \(t = \frac{ln(\frac{7500}{2500})}{0.0375} = \frac{ln(3)}{0.0375}.\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Continuous Compounding
When you hear the term continuous compounding, it refers to a way interest can be calculated and added to an investment's balance continuously, at every possible moment. Unlike regular compounding, where interest is compounded at intervals like annually, semi-annually, quarterly, or monthly, continuous compounding means your money is growing at a constantly accelerating rate鈥攍iterally all the time.

This type of compounding is often related to the mathematical constant \( e \), which is approximately equal to 2.71828. The formula used for continuous compounding is \( A = Pe^{rt} \), where:
  • \( A \) is the final amount after time \( t \)
  • \( P \) is the principal or initial amount
  • \( r \) is the annual interest rate (expressed as a decimal)
  • \( e \) is the base of the natural logarithm system
Understanding this formula can help you calculate how long it will take your investment to grow to a certain amount with continuous interest.
Compound Interest Formula
The compound interest formula is a crucial concept in finance. For continuous compounding, the formula \( A = Pe^{rt} \) is used. It is derived from the general compound interest formula but assumes that interest calculation occurs infinitely often.
  • The formula takes into account not only the interest earned on your principal but also interest earned on interest that has already been added to your account balance.
  • This can significantly increase your investment's growth over time.
  • In problems involving doubling or tripling of investments, you set \( A = 2P \) or \( A = 3P \), respectively, and then solve for \( t \).
Understanding this formula helps you predict how investments grow and allows you to make informed financial decisions.
Logarithms
Logarithms play an essential role in solving equations involving continuous compounding. They provide a way to reverse the process of exponentiation, making them perfect for finding the variable \( t \) in our compound interest formula.
  • In the context of our problem, natural logarithms (denoted as \( \ln \)) are particularly useful, because they are tied to the base \( e \) of the exponential function.
  • To find the time \( t \) required to double or triple an investment, logarithms are used to isolate \( t \) in the equation.
  • For example, to find the doubling time, you rearrange \( 5000 = 2500 e^{0.0375t} \) to \( e^{rt} = 2 \), then apply the natural logarithm yielding \( t = \frac{\ln(2)}{0.0375} \).
Logarithms help transform complex multiplicative relationships into simpler additive ones, making calculations more manageable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the acidity model given by \(\mathbf{p H}=-\log \left[\mathbf{H}^{+}\right],\) where acidity \((\mathbf{p H})\) is a measure of the hydrogen ion concentration \(\left[\mathbf{H}^{+}\right]\) (measured in moles of hydrogen per liter) of a solution. Apple juice has a pH of 2.9 and drinking water has a pH of \(8.0 .\) The hydrogen ion concentration of the apple juice is how many times the concentration of drinking water?

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$2 x \ln \left(\frac{1}{x}\right)-x=0$$

Use a graphing utility to graph the function. Be sure to use an appropriate viewing window. \(f(x)=\ln x+8\)

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$\ln (x+1)=2-\ln x$$

Home Mortgage \(A \$ 120,000\) home mortgage for 30 years at \(7 \frac{1}{2} \%\) has a monthly payment of \(\$ 839.06\) Part of the monthly payment covers the interest charge on the unpaid balance, and the remainder of the payment reduces the principal. The amount paid toward the interest is $$u=M-\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}$$ and the amount paid toward the reduction of the principal is $$v=\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}$$ In these formulas, \(P\) is the size of the mortgage, \(r\) is the interest rate, \(M\) is the monthly payment, and \(t\) is the time (in years). (a) Use a graphing utility to graph each function in the same viewing window. (The viewing window should show all 30 years of mortgage payments.) (b) In the early years of the mortgage, is the greater part of the monthly payment paid toward the interest or the principal? Approximate the time when the monthly payment is evenly divided between interest and principal reduction. (c) Repeat parts (a) and (b) for a repayment period of 20 years \((M=\$ 966.71) .\) What can you conclude?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.