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Chapter 12: Problem 272

What is the time required for one dollar to double when invested at the rate of \(5 \%\) per annum? Assume that interest is compounded continuously.

Short Answer

Expert verified
The time required for one dollar to double when invested at the rate of 5% per annum, under continuous compounding, is approximately \(13.86\) years.

Step by step solution

01

Identify the continuous compounding interest formula

The continuous compounding interest formula is given by: \[A = P * e^{rt}\] Where: A represents the final amount P is the principal or initial amount r is the annual interest rate t is the time (in years) e is the base of natural logarithm (approximately 2.718) In this problem, we want to find the time required for the investment to double, which means the final amount (A) must be twice the initial amount (P).
02

Set up the equation and substitute the values

Since we are looking for the time it takes for the investment to double, we will set A = 2P. Now, we can substitute the values into the formula: \[2P = P * e^{rt}\] We also know the annual interest rate, r, is 5% or 0.05. Our equation becomes: \[2P = P * e^{0.05t}\]
03

Solve the equation for t

To solve for t, we need to first isolate the term containing t. Divide each side of the equation by P: \[2 = e^{0.05t}\] Next, we will take the natural logarithm (ln) of both sides to get rid of the exponent. \[\ln(2) = \ln(e^{0.05t})\] Using the logarithm property, we can move the exponent to the front: \[\ln(2) = 0.05t * \ln(e)\] Since base e and logarithm e cancel each other, we get: \[\ln(2) = 0.05t\] Now, we can find the value of t by dividing both sides by 0.05: \[t = \frac{\ln(2)}{0.05}\]
04

Calculate the value of t

Using a calculator, we can compute the value of t: \[t = \frac{\ln(2)}{0.05} \approx 13.86\]
05

Interpret the result

The time required for one dollar to double when invested at the rate of 5% per annum, under continuous compounding, is approximately 13.86 years.

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Key Concepts

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

Exponential Growth
When we talk about exponential growth, we're looking at situations where quantities increase at a rate proportional to their current value. This kind of growth is common in finance, biology, and many other fields.

For example, in finance, when you invest money at a continuous compound interest rate, your investment grows exponentially over time. The formula for continuous compounding, as seen in the exercise, is expressed as \(A = Pe^{rt}\). The 鈥榚鈥� in this equation is Euler's number, a fundamental constant approximately equal to 2.718, which serves as the base for natural logarithms.
This formula tells us that the future value (A) of an investment (P) increases exponentially based on the constant rate of growth (r) over a given time (t). In our case, to determine how quickly an investment doubles, we apply this concept to see how an initial amount grows to twice its size through continuous compounding.
Natural Logarithm
The natural logarithm is a mathematical operation that is the inverse of exponentiation when considering the base of the natural logarithm, 'e'. It's represented by 鈥榣n鈥� and plays a crucial role in solving equations associated with exponential growth.

In the exercise above, we used the natural logarithm to solve for the variable 鈥榯鈥� in the equation \(2 = e^{0.05t}\), by applying the logarithmic property \(\ln(e^x) = x\ln(e)\). Since \(e\) is the base of the natural logarithm, \(\ln(e)\) equals 1, simplifying the equation to \(\ln(2) = 0.05t\). This step transforms the exponential equation into a linear one, which we can easily solve for 鈥榯鈥�.
By understanding the natural logarithm, you can solve numerous financial equations involving continuous growth, as it allows you to isolate a growth rate or time variable within an exponential function.
Time Value of Money
The time value of money is a core principle in finance that describes the increase in the amount of money over a given period due to potential interest earned. The concept is based on the idea that a specific amount of money today is worth more than the same amount in the future because of its potential earning capacity.

The continuous compounding formula, \(A = Pe^{rt}\), embodies the time value of money by showing how money can grow over time when interest is compounded continuously. As seen in the exercise, the principal (P) can grow to an amount (A) over time (t) with a constant interest rate (r).
A practical application of this concept is when planning investments or retirement savings. Understanding how money grows over time helps investors make informed decisions about saving and investing, and it's essential to comprehend this to grasp the fundamentals of personal finance and economics.

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Most popular questions from this chapter

The temperature inside a house is \(70^{\circ} \mathrm{F}\). A thermometer is taken outside after being inside the house for enough time for it to read \(70^{\circ}\). The outside air temperature is \(10^{\circ} \mathrm{F}\). After three minutes the thermometer reading is found to be \(25^{\circ} \mathrm{F}\). Find the reading on the thermometer as a function of time.

Derive a practical expression determining the barometric pressure as a function of the height, \(\mathrm{h}\), above the surface of the earth, for applications at low altitudes. The pressure at the surface, in pounds per square inch, is \(14.7\). At 4500 feet, the pressure drops to \(12.5 \mathrm{lb} / \mathrm{in}^{2}\). You will need to use the fundamental gas law $$ \mathrm{N}=\left\\{\left(\mathrm{k}_{1} \mathrm{p} \mathrm{v}\right) /(\mathrm{T})\right\\} $$ where \(\mathrm{N}\) is the number of molecules of the gas, \(\mathrm{p}\) is the pressure, \(\mathrm{v}\) is the volume of the gas, and \(\mathrm{T}\) is the absolute temperature. The constant \(\mathrm{k}_{1}\) depends only upon the units being used.Suspended \(\mathrm{Cable}\)

Set up the expression for the rate of a bimolecular reaction (a) when the two initial reactants are identical and (b) when two different reactants are used. For (a) find the concentration in terms of the initial concentration, the rate constant and time, assuming that the reactants combine in a one \(-\) to \(-\) one ratio. When in a reaction two molecules react to produce one or more resultants it is branded as a bimolecular or Second Order Reaction. State some examples of bimolecular reactions.

The rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. Half of the original number of radioactive nuclei have undergone disintegration in a period of 1500 years. 1\. What percentage of the original radioactive nuclei will remain after 4500 years? 2\. In how many years will only one - tenth of the original number remain?

A certain compound \(\mathrm{X}\) is formed by the combination of 2 parts of a chemical \(\mathrm{U}\) with 3 parts of chemical \(\mathrm{W}\). When certain amounts of \(\mathrm{U}\) and \(\mathrm{W}\) are placed together, the rate at which \(\mathrm{X}\) is produced is constantly proportional to the product of the amounts of \(\mathrm{U}\) and \(\mathrm{W}\) that are still present at that instant. Determine the amount of \(X\), that is produced in time \(t\), if 10 lb. of \(U\) and 8 lb. of \(W\) are placed together at \(\mathrm{t}=0\) and the amount of \(\mathrm{X}\) at \(\mathrm{t}=1\) is \(2 \mathrm{lb}\).Economics
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