/*! 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 19 Suppose that under continuous co... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 19

Suppose that under continuous compounding of interest, the effective rate is \(6 \%\) per annum. Compute the nominal rate.

Short Answer

Expert verified
The nominal rate is approximately 5.827% per annum.

Step by step solution

01

Understand the Relationship between Nominal and Effective Rates

Under continuous compounding, the effective rate and the nominal rate are related through the formula: \( E = e^r - 1 \), where \( E \) is the effective rate, and \( r \) is the nominal rate.
02

Substitute the Known Effective Rate into the Formula

We are given that the effective rate, \( E \), is 6%, or \( 0.06 \) in decimal form. Substitute this into the formula: \( 0.06 = e^r - 1 \).
03

Solve for the Nominal Rate

To find \( r \), rearrange the equation: \( e^r = 0.06 + 1 = 1.06 \). Taking the natural logarithm of both sides, we get \( r = \ln(1.06) \).
04

Calculate the Nominal Rate

Calculate the natural logarithm with a calculator: \( r \approx \ln(1.06) \approx 0.05827 \). Thus, the nominal rate \( r \) is approximately 5.827%.

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.

Effective Interest Rate
The effective interest rate is an important concept in finance, especially when it comes to comparing different investment or loan options. It essentially tells you the true rate at which your money grows over a year, accounting for compounding.
  • This rate takes into consideration all the compounding periods within the year, whether monthly, quarterly, or continuously.
  • When considering investments or loans described with a nominal percentage, the effective interest rate provides a clearer picture of their true cost or return.
  • If a nominal rate is compounded continuously, the effective rate can be found through the formula: \[E = e^r - 1\] where \(E\) is the effective rate, and \(r\) is the nominal rate.
Understanding the effective interest rate equips you to see beyond surface-level interest rates and gauge the actual financial impact of interest over time.
Nominal Interest Rate
The nominal interest rate is often referred to as the "stated" or "annual" interest rate. It is the interest rate stated on the loan or investment without taking compounding into account.
  • This rate is usually presented annually but does not account for how frequently compounding occurs, whether it's monthly, quarterly, or even continuously.
  • In the context of continuous compounding, we relate the nominal rate to the effective rate using the formula: \[E = e^r - 1\] where \(E\) is the effective interest rate and \(r\) is the nominal rate.
  • To solve for the nominal rate in the given problem, we rearranged the formula and calculated using the natural logarithm: \(r = \ln(1.06)\).
Knowing the nominal interest rate is a starting point when evaluating an investment, although it's crucial to understand its limitations as it does not give a full picture without considering the compounding frequency.
Natural Logarithm
The natural logarithm is a mathematical function that is key to solving equations involving continuous growth, exponential processes, and, in this context, continuous compounding of interest.
  • Represented as \(\ln\), the natural logarithm is the inverse of the exponential function \(e^x\).
  • In our problem, it was used to find the nominal rate from the effective interest rate by solving \(r = \ln(1.06)\).
  • This calculation effectively reverses the effect of exponentiation, allowing us to determine the rate that, when compounded continuously, would yield the effective interest rate.
  • The expression \(\ln\) is crucial for any scenario where the relationship involves natural exponentials, common in continuous growth models such as population dynamics and interest compounding.
The natural logarithm is a powerful tool in calculus and financial mathematics, enabling deeper analysis and understanding of growth scenarios.

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

Prove that \(\left(\log _{a} x\right) /\left(\log _{a b} x\right)=1+\log _{a} b\)

Use the half-life information to complete each table. (The formula \(\mathcal{N}=\mathcal{N}_{0} e^{k t}\) is not required.) (a) Uranium-228: half-life \(=550\) seconds$$\begin{array}{llllll}t \text { (seconds) } & 0 & 550 & 1100 & 1650 & 2200 \\\\\mathcal{N} \text { (grams) } & 8 & & & \\\\\hline\end{array}$$ (b) Uranium-238: half-life \(=4.9 \times 10^{9}\) years$$\begin{array}{lcccccc}\hline \multirow{2}{*}\begin{array}{l}\text { t (years) } \\\\\mathcal{N} \text { (grams) }\end{array} & \multicolumn{2}{c}0 \\\& \multicolumn{2}{c}10 & 5 & 2.5 & 1.25 & 0.625 \\\\\hline\end{array}$$

Use the following information on \(p H\) Chemists define pH by the formula pH \(=-\log _{10}\left[\mathrm{H}^{+}\right],\) where [H \(^{+}\) ] is the hydrogen ion concentration measured in moles per liter. For example, if \(\left[\mathrm{H}^{+}\right]=10^{-5},\) then \(p H=5 .\) Solutions with \(a\) pH of 7 are said to be neutral; a p \(H\) below 7 indicates an acid: and a pH above 7 indicates a base. (A calculator is helpful for Exercises 49 and 50.1 What is the hydrogen ion concentration for black coffee if the pH is \(5.9 ?\)

Let \(\mathcal{N}=\mathcal{N}_{0 e^{k t}} .\) In this exercise we show that if \(\Delta t\) is very small, then \(\Delta \mathcal{N} / \Delta t \approx k \mathcal{N} .\) In other words, over very small intervals of time, the average rate of change of \(\mathcal{N}\) is proportional to \(\mathcal{N}\) itself. (a) Show that the average rate of change of the function \(\mathcal{N}=\mathcal{N}_{0} e^{t t}\) on the interval \([t, t+\Delta t]\) is given by $$\frac{\Delta \mathcal{N}}{\Delta t}=\frac{\mathcal{N}_{0} e^{k t}\left(e^{k \Delta t}-1\right)}{\Delta t}=\frac{\mathcal{N}\left(e^{k \Delta t}-1\right)}{\Delta t}$$ (b) In Exercise 26 of Section 5.2 we saw that \(e^{x} \approx x+1\) when \(x\) is close to zero. Thus, if \(\Delta t\) is sufficiently small, we have \(e^{k \Delta t} \approx k \Delta t+1 .\) Use this approximation and the result in part (a) to show that \(\Delta \mathcal{N} / \Delta t \approx k N\) when \(\Delta t\) is sufficiently close to zero.

This exercise demonstrates the very slow growth of the natural logarithm function \(y=\ln x .\) We consider the following question: How large must \(x\) be before the graph of \(y=\ln x\) reaches a height of \(10 ?\) (a) Graph the function \(y=\ln x\) using a viewing rectangle that extends from 0 to 10 in the \(x\) -direction and 0 to 12 in the \(y\) -direction. Note how slowly the graph rises. Use the graphing utility to estimate the height of the curve (the \(y\) -coordinate) when \(x=10\) (b) since we are trying to see when the graph of \(y=\ln x\) reaches a height of 10 , add the horizontal line \(y=10\) to your picture. Next, adjust the viewing rectangle so that \(x\) extends from 0 to \(100 .\) Now use the graphing utility to estimate the height of the curve when \(x=100 .\) [As both the picture and the \(y\) -coordinate indicate, we're still not even halfway to \(10 .\) Go on to part (c).] (c) Change the vicwing rectangle so that \(x\) extends to \(1000,\) then estimate the \(y\) -coordinate corresponding to \(x=1000 .\) (You'll find that the height of the curve is almost \(7 .\) We're getting closer.) (d) Repeat part (c) with \(x\) extending to \(10,000 .\) (You'll find that the height of the curve is over \(9 .\) We're almost there. \()\) (e) The last step: Change the viewing rectangle so that \(x\) extends to \(100,000,\) then use the graphing utility to estimate the \(x\) -value for which \(\ln x=10 .\) As a check on your estimate, rewrite the equation \(\ln x=10\) in exponential form, and evaluate the expression that you obtain for \(x\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Mechanics Maths

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.