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Account Estimate how long it will take for \(\$ 5000\) to grow to \(\$ 8400\) at an interest rate of \(6 \%\) if interest is compounded (a) semiannually; (b) continuously.

Short Answer

Expert verified
8.51 years for semiannual; 8.61 years for continuous compounding.

Step by step solution

01

Understanding Compound Interest Formula (Semiannually)

To find out how long it will take for the money to grow to a specific amount with semiannual compounding, use the formula \( A = P \left(1 + \frac{r}{n}\right)^{nt} \), where \( A \) is the amount of money accumulated after \( t \) years, including interest. \( P \) is the principal amount (\$5000), \( r \) is the annual interest rate (0.06), \( n \) is the number of times that interest is compounded per year (2 for semiannual), and \( t \) is the time in years.
02

Plugging in Values (Semiannually)

Set the equation with \( A = 8400 \), \( P = 5000 \), \( r = 0.06 \), and \( n = 2 \). The formula becomes \[ 8400 = 5000 \, \left(1 + \frac{0.06}{2}\right)^{2t} \].
03

Solving for Time (Semiannually)

First, simplify the term inside the parentheses: \( 1 + \frac{0.06}{2} = 1.03 \). Next, divide both sides of the equation by 5000 to isolate the term with \( t \): \( \frac{8400}{5000} = 1.03^{2t} \). This yields \( 1.68 = 1.03^{2t} \). Taking the natural logarithm of both sides, we have \( \ln(1.68) = 2t \cdot \ln(1.03) \). Solve for \( t \) by dividing both sides by \( 2 \cdot \ln(1.03) \).
04

Final Calculation (Semiannually)

Calculate \( t \) using the equation \( t = \frac{\ln(1.68)}{2 \cdot \ln(1.03)} \). This gives us \( t \approx 8.51 \) years.
05

Understanding Continuous Compounding Formula

With continuous compounding, we use the formula \( A = Pe^{rt} \), where \( e \) is Euler's number.
06

Plugging in Values (Continuously)

Set the formula with \( A = 8400 \), \( P = 5000 \), \( r = 0.06 \). The equation becomes \[ 8400 = 5000e^{0.06t} \].
07

Solving for Time (Continuously)

First, divide both sides by 5000: \( \frac{8400}{5000} = e^{0.06t} \), simplifying to \( 1.68 = e^{0.06t} \). Take the natural logarithm of both sides: \( \ln(1.68) = 0.06t \). Solve for \( t \) by dividing both sides by 0.06.
08

Final Calculation (Continuously)

The equation becomes \( t = \frac{\ln(1.68)}{0.06} \). This calculation results in \( t \approx 8.61 \) years.

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Key Concepts

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

Continuous Compounding
Continuous compounding is a unique way of calculating interest that assumes the principal amount grows continuously. In simpler terms, it's like having your money working for you all the time without any breaks. The formula used for continuous compounding is:\[ A = Pe^{rt} \]
  • **A** is the amount of money including interest after time **t**.
  • **P** is the principal or the starting amount (e.g., \(5000).
  • **r** is the annual interest rate (for example, 0.06 for 6%).
  • **t** is the time in years.
  • **e** is a mathematical constant approximately equal to 2.718, known as Euler's number.
To find out how long it takes for an investment to grow to a desired amount using continuous compounding, rearrange the formula to solve for **t**. This involves dividing both sides by **P**, then taking the natural logarithm (ln) of both sides to isolate **t**. In our example, it takes about 8.61 years for \)5000 to grow to $8400 when compounded continuously at a 6% annual rate. This is because the interest is constantly being calculated and added, allowing it to accumulate more quickly.
Semiannual Compounding
In semiannual compounding, interest is calculated and added to the principal twice a year. This means every six months, the accumulated interest is reinvested to earn more interest. The formula used for semiannual compounding is:\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
  • **A** is the total amount after interest.
  • **P** is the initial amount of \(5000.
  • **r** is the annual interest rate, 0.06.
  • **n** is the number of compounding periods per year, which is 2 for semiannual.
  • **t** is the time in years.
Using this formula, you substitute the values into the equation and solve for **t**. For our example with semiannual compounding, it takes about 8.51 years for the principal of \)5000 to grow to $8400. This process involves calculating the effective interest rate, dividing this by the number of compounding periods, and then using logarithms to solve for the time required.
Exponential Growth
Exponential growth describes a process where the quantity increases at a rate proportional to its current value. This concept often appears in finance when dealing with compounded interest. In our example, both continuous and semiannual compounding reflect exponential growth, as they illustrate how money can multiply over time when interest is continually added. Some features of exponential growth in financial contexts are:
  • Growth becomes faster as the principal amount increases because interest is calculated on the previously accumulated interest.
  • Small increases in the interest rate or compounding frequency can significantly affect the growth rate.
For instance, compounding interest continuously will grow an investment slightly faster than semiannual compounding due to the ongoing accumulation of interest. Understanding these effects is critical for financial planning, allowing one to estimate how different levels of interest rates and compounding frequencies will impact investment outcomes over time.

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Most popular questions from this chapter

Escherichia coli is a strain of bacteria that occurs naturally in many organisms. Under certain conditions, the number of bacteria present in a colony is approximated by $$A(t)=A_{0} e^{0.023 t}$$ where \(t\) is in minutes. If \(A_{0}=2,400,000,\) find the number of bacteria at each time. Round to the nearest hundred thousand. (a) 5 minutes (b) 10 minutes (c) 60 minutes

Estimate the doubling time of an investment earning \(2.5 \%\) interest if interest is compounded (a) quarterly; (b) continuously.

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Use the table feature of your graphing calculator to work parts (a) and (b). (a) Find how long it will take \(\$ 1500\) invested at \(5.75 \%\) compounded daily, to triple in value. Locate the solution by systematically decreasing \(\Delta\) Tbl. Find the answer to the nearest day. (Find your answer to the nearest day by eventually letting \(\Delta \mathrm{Tbl}=\frac{1}{365} .\) The decimal part of the solution can be multiplied by 365 to determine the number of days greater than the nearest year. For example, if the solution is determined to be 16.2027 years, then multiply 0.2027 by 365 to get \(73.9855 .\) The solution is then, to the nearest day, 16 years and 74 days.) Confirm your answer analytically. (b) Find how long it will take \(\$ 2000\) invested at \(8 \%,\) compounded daily, to be worth \(\$ 5000\).

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