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Chapter 5: Problem 14

A principal of \(\$ 4000\) dollars is invested at \(8 \%\) per annum compounded continuously. (a) Use a graphing utility to estimate how long it will take for the balance to increase by \(25 \% .\) (That is, you want a balance of \(\$ 4000+\$ 1000=\$ 5000 .)\) (Adapt the suggestion at the end of Example \(5 .\) ) (b) Use algebra, rather than a graphing utility, to solve the problem in part (a).

Short Answer

Expert verified
The investment will take about 2.789 years to reach $5000.

Step by step solution

01

Understanding the Problem

We want to find out how many years it takes for a $4000 investment compounded continuously at 8% per annum to reach $5000.
02

Understanding Continuous Compounding Formula

The continuous compounding formula is given by \( A = Pe^{rt} \), where \( A \) is the amount after time \( t \), \( P \) is the principal amount, \( r \) is the rate of interest, and \( t \) is the time in years. We're given \( P = 4000 \), \( r = 0.08 \), and \( A = 5000 \).
03

Setting Equation for Time

Plug the known values into the formula: \[ 5000 = 4000e^{0.08t}. \] We need to solve for \( t \).
04

Simplifying the Equation

Divide both sides by 4000 to isolate the exponential term: \[ \frac{5000}{4000} = e^{0.08t} \Rightarrow 1.25 = e^{0.08t}. \]
05

Solving for t - Taking Natural Logarithm

Take the natural logarithm of both sides to solve for \( t \): \[ \ln(1.25) = 0.08t. \]
06

Isolating t

Divide by 0.08 to solve for \( t \): \[ t = \frac{\ln(1.25)}{0.08}. \]
07

Calculating t

Plug the values into a calculator to find \( t \): \[ t \approx \frac{0.2231}{0.08} \approx 2.789 \text{ years}. \]
08

Final Answer

The time required for the investment to reach $5000 is approximately 2.789 years.

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Key Concepts

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

Compound Interest
Compound interest is a powerful concept in finance that allows money to grow over time. It occurs when the interest earned on an investment is reinvested, leading to interest being earned on top of interest. Here's how it works:

  • **Principal:** The original sum of money invested.
  • **Interest Rate:** The percentage at which the investment grows annually.
  • **Compounding Frequency:** The number of times interest is calculated and added to the principal in a given timeframe.
Continuous compounding is a special case where the compounding frequency is theoretically infinite. This means interests are added at every possible moment. This is described by the formula \( A = Pe^{rt} \), where \( A \) is the amount of money accumulated after \( t \) years, \( P \) is the principal, \( r \) is the annual interest rate, and \( t \) is the time.

Understanding compound interest helps you evaluate financial decisions and the growth potential of your investments over time. Whether you are a student or someone planning their financial future, grasping this concept is crucial.
Exponential Growth
Exponential growth describes a process where the quantity increases at a rate proportional to its current value. This concept is widely used in natural sciences and finance. With exponential growth, a larger amount results in a larger increase, leading to rapid expansion over time.

When investing money, exponential growth explains how small amounts can significantly grow over long periods through compounding. Using the formula connected to continuous compounding \( A = Pe^{rt} \), we can see that the expression \( e^{rt} \) represents the exponential growth rate.

Exponential growth is different from linear growth, where increases are constant over time. With exponential growth, changes happen faster and reach higher values quickly. This makes it especially relevant in understanding investments, populations, and other naturally growing phenomena. Through continuous compounding, you can see tangible examples of exponential growth in action.
Natural Logarithm
The natural logarithm (\( \ln \)) is a mathematical function often used to solve problems involving exponential growth and decay. It is the inverse of the exponential function, meaning it helps us find the exponent that would produce a given number.

In the formula for continuous compounding, \( A = Pe^{rt} \), we use the natural logarithm to solve for time \( t \). If we need to find out how long it takes for an investment to grow to a certain amount, we rearrange the equation to take the form \( e^{rt} = \text{some value} \), and then apply the natural logarithm:
\( \ln(A/P) = rt \).

The natural logarithm has unique properties that make it particularly useful:
  • \( \ln(e^x) = x \): This property states that the natural log of an exponentiated \( e \) value returns the original exponent \( x \).
  • \( \ln(1) = 0 \): Because \( e^0 = 1 \), this makes solving equations convenient.
When working with continuous compounding and growth processes, the natural logarithm is essential for breaking down and solving equations for unknown time factors.

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Most popular questions from this chapter

A sound level of \(\beta=120 \mathrm{db}\) is at the threshold of pain. (Some loud rock concerts reach this level.) The sound intensity that corresponds to \(\beta=120 \mathrm{db}\) is \(1 \mathrm{W} / \mathrm{m}^{2}\). Use this information and the equation \(\beta=10 \log _{10}\left(I / I_{0}\right)\) to determine \(I_{0}\), the intensity of a barely audible sound at the threshold of hearing. What is the decibel level, \(\beta\), of a barely audible sound?

The age of some rocks can be estimated by measuring the ratio of the amounts of certain chemical elements within the rock. The method known as the rubidium-strontium method will be discussed here. This method has been used in dating the moon rocks brought back on the Apollo missions. Rubidium-87 is a radioactive substance with a half-life of \(4.7 \times 10^{10}\) years. Rubidium- 87 decays into the substance strontium- \(87,\) which is stable (nonradioactive). We are going to derive the following formula for the age of a rock: $$T=\frac{\ln \left[\left(\mathcal{N}_{s} / \mathcal{N}_{r}\right)+1\right]}{-k}$$ where \(T\) is the age of the rock, \(k\) is the decay constant for rubidium-87, \(\mathcal{N}_{s}\) is the number of atoms of strontium-87 now present in the rock, and \(\mathcal{N},\) is the number of atoms of rubidium-87 now present in the rock. (a) Assume that initially, when the rock was formed, there were \(\mathcal{N}_{0}\) atoms of rubidium-87 and none of strontium-87. Then, as time goes by, some of the rubidium atoms decay into strontium atoms, but the to tal number of atoms must still be \(\mathcal{N}_{0} .\) Thus, after \(T\) years, we have \(\mathcal{N}_{0}=\mathcal{N}_{r}+\mathcal{N}_{s}\) or, equivalently, $$ \mathcal{N}_{s}=\mathcal{N}_{0}-\mathcal{N}_{r}$$However, according to the law of exponential decay for the rubidium-87, we must have \(\mathcal{N}_{r}=\mathcal{N}_{0} e^{k T} .\) Solve this equation for \(\mathcal{N}_{0}\) and then use the result to eliminate \(\mathcal{N}_{0}\) from equation \((1) .\) Show that the result can be written $$\mathcal{N}_{s}=\mathcal{N}_{r} e^{-k T}-\mathcal{N}_{r}$$ (b) Solve equation (2) for \(T\) to obtain the formula given at the beginning of this exercise.

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$\ln \frac{3 x-2}{4 x+1}>\ln 4$$

The intensity of the sounds that the human ear can detect varies over a very wide range of values. For instance, a whisper from 1 meter away has an intensity of approximately \(10^{-10}\) watts per square meter \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\), whereas, from a distance of 50 meters, the intensity of a launch of the Space Shuttle is approximately \(10^{8} \mathrm{W} / \mathrm{m}^{2} .\) For a sound with intensity \(I\), the sound level \(\beta\) is defined by $$ \beta=10 \log _{10}\left(I / I_{0}\right) $$ where the constant \(I_{0}\) is the sound intensity of a barely audible sound at the threshold of hearing. The units for the sound level \(\beta\) are decibels, abbreviated dB. (a) Solve the equation \(\beta=10 \log _{10}\left(1 / I_{0}\right)\) for \(I\) by first dividing by 10 and then converting to exponential form. (b) The sound level for a power lawnmower is \(\beta=100 \mathrm{db}\). and that for a cat purring is \(\beta=10\) db. Use your result in part (a) to determine how many times more intense is the power mower sound than the cat's purring.

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$e^{1 /(x-1)}>1$$
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