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Chapter 3: Problem 71

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. $$r=0.025$$

Short Answer

Expert verified
The time required for the investment to double is approximately \( \frac{\ln{2}}{0.025} \) years, and for the investment to triple it is about \( \frac{\ln{3}}{0.025} \) years. So the answer for (a) and (b) will be numerical values for these expressions respectively.

Step by step solution

01

Calculating Time to Double Investment

For the amount to double, \(A\) will be \(2 * 2500 = \$5000\). Now using the formula of continuous compounding \(A = Pe^{rt}\), we can substitute the given and derived values to the equation: \[5000 = 2500 * e^{(0.025t)}\] Further we simplify this equation and solve for \(t\): \[2 = e^{(0.025t)}\] Now, to get \(t\), we take natural logarithm on both sides: \[\ln{2} = 0.025t\] Hence, \(t = \frac{\ln{2}}{0.025}\)
02

Calculating Time to Triple Investment

To calculate the time required for the investment to triple, \(A\) will be \(3 * 2500 = \$7500\). We again use the formula of continuous compounding \(A = Pe^{rt}\), and substitute the given and derived values: \[7500 = 2500 * e^{(0.025t)}\] After simplifying the equation, we get: \[3 = e^{(0.025t)}\] We take natural logarithm on both sides to get \(t\): \[\ln{3} = 0.025t\] Consequently, \(t = \frac{\ln{3}}{0.025}\]

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Key Concepts

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

Exponential Growth
Exponential growth describes a process where the growth rate of a mathematical function is proportional to the function's current value. In finance, this concept typically applies to the growth of investments and interest. Key to this is the principle of an investment increasing at a rate that is constant over time, leading to a steeper and steeper increase in the investment's total value as time progresses.

When investments compound continuously, as in the investment exercise we see, this is an ideal example of exponential growth. The formula representing continuous compounding is given by A = Pe^{rt}, where A is the amount of money after a certain time t, P is the principal amount (initial investment), r is the annual interest rate, and e is the base of the natural logarithm, approximately equal to 2.71828.

The exercise shows that with a continuous interest rate, the original investment grows exponentially over time, meaning it doesn't just increase by a fixed amount each year but rather that increase becomes larger as time goes on, provided the rate r remains constant.
Natural Logarithm
The natural logarithm, denoted as ln, is the logarithm to the base e, where e is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm is particularly useful when dealing with exponential growth, as it can be used to reverse the effects of exponentiation with the base e.

In our exercise, natural logarithm plays a critical role. When we have the equation 2 = e^{(0.025t)} or 3 = e^{(0.025t)}, to solve for t, we need to get rid of the exponential term e. We do this by taking the natural logarithm of both sides, which effectively 'unwraps' the exponent, allowing us to solve for t.

After applying the natural logarithm, the equations reduce to ln(2) = 0.025t and ln(3) = 0.025t. We can then easily isolate t to find the time it takes for the initial investment to double and triple, demonstrating how logarithms are a key tool in understanding and working with exponential functions.
Time Value of Money
The time value of money is a finance concept that reflects the idea of the opportunity cost of not having money at the present time. It incorporates the belief that money available today is worth more than the same amount in the future, due to its potential earning capacity.

This principle is one of the core reasons for interest, which compensates for the time an investor must wait to access their money. Compounded interest, especially continuous compounding as shown in the given exercise, shows how money's value changes over time. With continuous compounding, small amounts contributed at regular intervals can significantly increase an investment's future value by accumulating earnings continuously.

As demonstrated in our exercise, an investor would need to understand the time value of money to make prudent decisions about when and how much to invest. By comprehending the mechanics behind continuous compounding, one can calculate the precise time it takes for their investment to double or triple, illustrating the practical utility of this concept in real-life financial planning and investment strategies.

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Most popular questions from this chapter

Exercises \(7-10\), rewrite the logarithm as a ratio of (a) common logarithms and (b) natural logarithms. $$\log _{5} 16$$

A cup of water at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C} .\) The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form \((t, T),\) where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). \(\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)\) \(\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right)\) (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \((t, T)\) and \((t, T-21)\) (b) An exponential model for the data \((t, T-21)\) is given by \(T-21=54.4(0.964)^{t} .\) Solve for \(T\) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use the graphing utility to plot the points \((t, \ln (T-21))\) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form \(\ln (T-21)=a t+b\) Solve for \(T,\) and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \(y\) -coordinates of the revised data points to generate the points $$\left(t, \frac{1}{T-21}\right)$$ Use the graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. The resulting line has the form $$\frac{1}{T-21}=a t+b$$ (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

Graphical Analysis Use a graphing utility to graph \(f\) and \(g\) in the same viewing window and determine which is increasing at the greater rate as \(x\) approaches + \(\infty\). What can you conclude about the rate of growth of the natural logarithmic function? (a) \(f(x)=\ln x, \quad g(x)=\sqrt{x}\) (b) \(f(x)=\ln x, \quad g(x)=\sqrt[4]{x}\)

The graph of \(f(x)=\log _{3} x\) contains the point \((27,3)\)

Forensics At 8: 30 A.M., a coroner went to the home of a person who had died during the night. In order to estimate the time of death, the coroner took the person's temperature twice. At 9: 00 A.M. the temperature was \(85.7^{\circ} \mathrm{F},\) and at 11: 00 A.M. the temperature was \(82.8^{\circ} \mathrm{F}\). From these two temperatures, the coroner was able to determine that the time elapsed since death and the body temperature were related by the formula $$t=-10 \ln \frac{T-70}{98.6-70}$$ where \(t\) is the time in hours elapsed since the person died and \(T\) is the temperature (in degrees Fahrenheit) of the person's body. (This formula comes from a general cooling principle called Newton's Law of Cooling. It uses the assumptions that the person had a normal body temperature of \(98.6^{\circ} \mathrm{F}\) at death and that the room temperature was a constant \(70^{\circ} \mathrm{F}\).) Use the formula to estimate the time of death of the person.
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