/*! 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 20 A bank offers \(6 \%\) compounde... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 20

A bank offers \(6 \%\) compounded continuously. How soon will a deposit: a. quadruple? b. increase by \(75 \% ?\)

Short Answer

Expert verified
a) About 23.11 years; b) About 9.33 years.

Step by step solution

01

Understand Continuous Compounding Formula

The formula for continuous compounding is given by \( A = Pe^{rt} \), where \( A \) is the amount of money accumulated after time \( t \), \( P \) is the principal amount (initial deposit), \( r \) is the annual interest rate (as a decimal), and \( e \) is the base of the natural logarithm (approximately 2.71828).
02

Set up the equation for a) quadrupling

To find when the deposit will quadruple, the amount \( A \) will be four times the principal \( P \). Set \( A = 4P \). Thus, we have: \[ 4P = Pe^{0.06t} \] We can simplify this to: \[ 4 = e^{0.06t} \]
03

Solve for time \( t \) in part a)

Take the natural logarithm on both sides of the equation: \[ \ln(4) = \ln(e^{0.06t}) \] This simplifies to: \[ \ln(4) = 0.06t \] Now, solve for \( t \) by dividing both sides by 0.06: \[ t = \frac{\ln(4)}{0.06} \] Calculate \( t \) using a calculator. \( \ln(4) \approx 1.3863 \), so \[ t \approx \frac{1.3863}{0.06} \approx 23.105 \text{ years} \]
04

Set up the equation for b) increasing by 75%

A 75% increase means the final amount \( A \) is 1.75 times the original deposit \( P \). Thus, set \( A = 1.75P \). Hence, \[ 1.75P = Pe^{0.06t} \] Simplify to get: \[ 1.75 = e^{0.06t} \]
05

Solve for time \( t \) in part b)

Take the natural logarithm of both sides: \[ \ln(1.75) = \ln(e^{0.06t}) \] This simplifies to: \[ \ln(1.75) = 0.06t \] Dividing both sides by 0.06 gives: \[ t = \frac{\ln(1.75)}{0.06} \] Calculate \( t \). \( \ln(1.75) \approx 0.55962 \), so \[ t \approx \frac{0.55962}{0.06} \approx 9.327 \text{ years} \]

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.

Exponential Growth
Exponential growth is a process that increases quantity over time. In this type of growth, the rate of increase is proportional to the current amount. This means the larger the amount becomes, the faster it grows. A common example of this concept is the way money grows in a bank account with interest. When an investment grows at a continuous rate, it follows an exponential growth pattern.
This continuous increase can be modeled mathematically by the equation: \[ A = Pe^{rt} \] where:
  • \( A \) is the final amount.
  • \( P \) is the principal or initial amount.
  • \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
  • \( r \) is the rate of growth (annual interest rate).
  • \( t \) represents time, typically in years.
This formula allows us to determine how investments grow under continuous compounding, a situation where interest is constantly being added to the principal balance.
Natural Logarithms
Natural logarithms are logarithms to the base \( e \). Written as \( \ln(x) \), they are fundamental in calculus and are particularly useful in solving equations involving exponential growth or decay.
In the context of continuous compounding, natural logarithms help isolate the variable of time \( t \) in the formula \( A = Pe^{rt} \).
Natural logarithms have a special property where the logarithm of an exponential function simplifies the equation significantly. For example, if \( y = e^{x} \), then \( \ln(y) = x \). This property allows us to solve for \( t \) in the equation by taking the natural log on both sides, transforming the nonlinear equation into a linear form that is much easier to solve.
Interest Rate Calculations
Calculating interest rates correctly is crucial for understanding how investments grow over time, especially with continuous compounding. If an interest rate is given as a percentage, it should be converted into a decimal for mathematical calculations. For example, an interest rate of 6% becomes 0.06.
In problems of continuous compounding, the key is to determine how long it takes for an amount to increase to a desired level. This is done by setting the exponential growth equation equal to the target amount and solving for time. Utilizing the natural logarithm helps in isolating the variable \( t \) when solving:
  • To determine when an amount quadruples, set \( A = 4P \), leading to \( 4 = e^{0.06t} \).
  • If an amount needs to increase by 75%, set \( A = 1.75P \), leading to \( 1.75 = e^{0.06t} \).
Through these calculations, one can quickly find how long it takes for investments to grow at a particular interest rate.
Financial Mathematics
Financial mathematics includes a range of topics that help us understand economic systems' quantitative and practical aspects. Interest, specifically compounded continuously, is an essential concept in financial mathematics. It shows how time and interest rates affect investments.
When using continuous compounding, the power of exponential growth fully utilizes the potential of an investment. By understanding the math behind it, investors can make informed decisions about their finances.
For practical applications, such calculations can guide:
  • Banking decisions: Choosing between different banks and interest rates.
  • Investment strategies: Understanding when an investment will meet a particular growth goal.
  • Budgeting: Planning ahead for significant financial landmarks.
By mastering these concepts, one can better manage personal or business finances, making the most of the financial opportunities available.

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

If an amount is invested at interest rate \(r\) compounded continuously, the doubling time (the time in which it will double in value) is found by solving the equation \(P e^{r t}=2 P\). The solution (by the usual method of canceling the \(P\) and taking logs) is \(t=\frac{\ln 2}{r} \approx \frac{0.69}{r} .\) For annual compounding, the doubling time should be somewhat longer, and may be estimated by replacing 69 by 72 . For example, to estimate the doubling time for an investment at \(8 \%\) compounded annually we would divide 72 by \(8,\) giving \(\frac{72}{8}=9\) years. The \(72,\) however, is only a rough "upward adjustment" of \(69,\) and the rule is most accurate for interest rates around \(9 \% .\) For each interest rate: a. Use the rule of 72 to estimate the doubling time for annual compounding. b. Use the compound interest formula \(P(1+r)^{t}\) to find the actual doubling time for annual compounding. $$ 1 \% $$

BIOMEDICAL: DWI and Crash Risk The crash risk of an intoxicated driver relative to a similar driver with zero blood alcohol is $$ R(x)=\frac{195}{1+1650 e^{-35.5 x}} $$ where \(x\) is the blood alcohol level as a percent \((0.01 \leq x \leq 0.25) .\) For example, \(R(0.16)=29.4\) means that a driver with blood alcohol level \(0.16 \%\) is 29.4 times more likely to be involved in an accident than a similar driver who is not impaired. Find \(R^{\prime}(0.16)\) and interpret your answer. [Note: \(x \geq 0.08\) defines "driving while intoxicated."]

101-107. Choose the correct answer: $$ \begin{aligned} &\frac{d}{d x} \ln f(x)=\text { a. } \frac{f(x)}{f^{\prime}(x)} \quad \text { b. } \frac{f^{\prime}(x)}{f(x)}\\\ &\text { c. } \frac{f^{\prime}(x)}{f^{\prime}(x)} \end{aligned} $$

69-72. Use your graphing calculator to graph each function on a window that includes all relative extreme points and inflection points, and give the coordinates of these points (rounded to two decimal places). [Hint: Use NDERIV once or twice with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=e^{x}+e^{-x} $$

Suppose that you have a bank account with interest compounded continuously, but you can't remember the continuously compounded interest rate. If at the end of the year you had \(10 \%\) more than you began with, was the continuously compounded rate more than \(10 \%\) or less than \(10 \% ?\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

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.