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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. $$r=0.0375$$

Short Answer

Expert verified
Therefore, it will take approximately \( ln(2) / 0.0375 \approx 18.47 \) years for the amount to double and \( ln(3) / 0.0375 \approx 29.61 \) years for the amount to triple.

Step by step solution

01

Analyze and Identify

Recognize that you can use the formula for continuous compounding, \(A = P * e^{rt}\), and you need to solve for \(t\) when \(A/P\) is 2 and then 3. The rate, \(r\), is given at 0.0375 or 3.75%.
02

Solve for the Doubling Time

Plug \(A/P = 2\), and \(r = 0.0375\) into the formula and solve for \(t\). This gives the equation \(2 = e^{0.0375t}\). To solve for \(t\), you'll need to use natural logarithms (ln). So, \(t = ln(2) / 0.0375\). Calculate this to get the approximate value for \(t\).
03

Solve for the Tripling Time

Similar to step 2, this time plug \(A/P = 3\), and \(r = 0.0375\) into the formula and solve for \(t\). This would give you the equation \(3 = e^{0.0375t}\). Similar to step 2, solve for \(t\) with natural logarithms (ln), \(t = ln(3) / 0.0375\). Calculate this to get the approximate value for \(t\).

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Key Concepts

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

Compound Interest Formula
Understanding the compound interest formula is crucial for studying the way investments grow over time. In contrast to simple interest, which is calculated only on the principal amount, compound interest means that any interest earned also earns interest in future periods. This leads to exponential growth of your investment.

For continuous compounding, the formula you'll often encounter is:
\[ A = Pe^{rt} \]
Here, \( A \) is the final amount of money after interest, \( P \) is the principal amount (initial investment), \( r \) is the annual interest rate (expressed as a decimal), \( t \) is the time the money is invested for, and \( e \) is Euler's number (approximately equal to 2.71828). This formula is powerful because it assumes that your investment is earning interest an infinite number of times per year, essentially continuously. It's an ideal model for situations like constantly fluctuating stock market investments or bank accounts that compound interest daily.
Natural Logarithm
When dealing with continuous compounding and exponential growth, the natural logarithm (ln) becomes our mathematical ally. The natural logarithm is the inverse operation to the exponential function with base \( e \), Euler's number. This means that, for any positive number \( x \), if you have \( e^y = x \), then \( y = ln(x) \).

Why is this useful? When you're solving for time \( t \) in the compound interest formula and you have an equation like \( 2 = e^{0.0375t} \), the natural logarithm helps you isolate \( t \). Here's how it works:
\[ ln(2) = ln(e^{0.0375t}) \]
Using the property that \( ln(e^x) = x \), you simplify it to:
\[ ln(2) = 0.0375t \]
Then, by dividing both sides by the interest rate \( r \), you get the value for \( t \). This process is essential in calculating exact times for investments to reach certain milestones, like doubling or tripling, under continuous compounding.
Exponential Growth
Exponential growth is a pattern of data that shows greater increases over time, creating a curve on a graph that becomes steeper and steeper. It's not just relevant in finance, but also in population growth, radioactive decay, and more. Importantly, exponential growth is characterized by the rate of growth becoming faster as time goes on.

In the context of continuous compounding, money doesn't just grow; it grows at an increasing rate because interest is constantly being calculated and added to the principal. This compounding effect causes wealth to expand at an exponential rate, which can be very powerful over longer periods. The key takeaway: the earlier you start investing, the more significant the benefits of exponential growth will be, thanks to the time value of money.
Time Value of Money
The time value of money is a financial concept that represents the idea that money available right now is worth more than the same amount in the future due to its potential earning capacity. This core principle of finance holds that, provided money can earn interest, any amount of money is worth more the sooner it is received.

When you invest with continuous compounding, the time value of money comes into clear focus. By investing earlier, you gain more interest, and that interest earns interest as well—this is the foundation of growing wealth. For example, given the same interest rate, an investment will grow much more if it starts with a higher principal or if it is given more time to grow. This is why savvy investors pay close attention to interest rates and start investing as early as possible.

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Most popular questions from this chapter

Condense the expression to the logarithm of a single quantity. $$\frac{1}{4} \log _{3} 5 x$$

Find a logarithmic equation that relates \(y\) and \(x .\) Explain the steps used to find the equation. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 1 & 1.189 & 1.316 & 1.414 & 1.495 & 1.565 \\ \hline\end{array}$$

Use the properties of logarithms to expand the expression as a sum, difference, and or constant multiple of logarithms. (Assume all variables are positive.) $$\log 4 x^{2} y$$

Engineers design automobiles with crumple zones that help protect their occupants in crashes. The crumple zones allow the occupants to move short distances when the automobiles come to abrupt stops. The greater the distance moved, the fewer \(\mathrm{g}\) 's the crash victims experience. (One \(\mathrm{g}\) is equal to the acceleration due to gravity.) In crash tests with vehicles moving at 90 kilometers per hour, analysts measured the numbers of g's experienced during deceleration by crash dummies that were permitted to move \(x\) meters during impact. The table shows the data. A model for the data is given by \(y=-3.00+11.88 \ln x+(36.94 / x),\) where \(y\) is the number of g's. $$\begin{array}{|l|l|l|l|l|l|}\hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\\\\hline \text { g's } & 158 & 80 & 53 & 40 & 32 \\\\\hline\end{array}$$ (a) Complete the table using the model.$$\begin{array}{|l|l|l|l|l|l|}\hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\\\\hline y & & & & & \\\\\hline\end{array}$$ (b) Use a graphing utility to graph the data points and the model in the same viewing window. How do they compare? (c) Use the model to estimate the distance traveled during impact, assuming that the passenger deceleration must not exceed \(30 \mathrm{g}\) 's. (d) Do you think it is practical to lower the number of g's experienced during impact to fewer than \(23 ?\) Explain your reasoning.

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\log 8 x-\log (1+\sqrt{x})=2$$

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