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We have to deal with problem-solving in many real-world situations. Therefore, it is important to know the steps you must take when problem-solving depending on the type of problem. Let's use exponential functions to solve the following problems:Suppose $4000$4000 is invested at a 6%6% interest rate compounded annually. How much money will there be in the bank at the end of five years? At the end of 2020 years?Read the problem and summarize the information.$$ is invested at a %% interest rate compounded annually. We want to know how much money we will have after and after years.Assign variables.x=𝑥= time in yearsy=𝑦= amount of money in the investment account.We start with $4000$4000 and each year we apply a 6%6% interest rate on the amount in the bank. The pattern is that each year we multiply the previous amount by a factor of 100%+6%=106%=1.06100%+6%=106%=1.06. Complete a table of values by continuing to multiply each year’s amount by 1.061.06.

Question

We have to deal with problem-solving in many real-world situations. Therefore, it is important to know the steps you must take when problem-solving depending on the type of problem. Let's use exponential functions to solve the following problems:Suppose 400040004000 is invested at a 6%6% interest rate compounded annually. How much money will there be in the bank at the end of five years? At the end of 2020 years?Read the problem and summarize the information. is invested at a %% interest rate compounded annually. We want to know how much money we will have after and after years.Assign variables.x=𝑥= time in yearsy=𝑦= amount of money in the investment account.We start with 400040004000 and each year we apply a 6%6% interest rate on the amount in the bank. The pattern is that each year we multiply the previous amount by a factor of 100%+6%=106%=1.06100%+6%=106%=1.06. Complete a table of values by continuing to multiply each year’s amount by 1.061.06.

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Solution

The problem is asking us to find out how much money will be in the bank after 5 years and after 2020 years, given an initial investment of $4000 and an annual interest rate of 6%.

Here are the steps to solve this problem:

  1. Identify the variables:

    • x = time in years
    • y = amount of money in the investment account

  2. Understand the pattern:

    • We start with $4000 and each year we apply a 6% interest rate on the amount in the bank.
    • The pattern is that each year we multiply the previous amount by a factor of 100%+6%=106%=1.06.

  3. Create a table of values:

    • To do this, we continue to multiply each year’s amount by 1.06.

Let's calculate:

For 5 years:

  • Year 1: 4000∗1.06=4000 * 1.06 = 4240
  • Year 2: 4240∗1.06=4240 * 1.06 = 4490.40
  • Year 3: 4490.40∗1.06=4490.40 * 1.06 = 4759.82
  • Year 4: 4759.82∗1.06=4759.82 * 1.06 = 5043.41
  • Year 5: 5043.41∗1.06=5043.41 * 1.06 = 5346.02

So, at the end of 5 years, there will be approximately $5346.02 in the bank.

For 2020 years, it would be impractical to calculate year by year. Instead, we can use the formula for compound interest, which is A = P(1 + r/n)^(nt), where:

  • A is the amount of money accumulated after n years, including interest.
  • P is the principal amount (the initial amount of money).
  • r is the annual interest rate (in decimal).
  • n is the number of times that interest is compounded per year.
  • t is the time the money is invested for in years.

In this case, our interest is compounded annually, so n = 1. Our rate is 6%, or 0.06 in decimal form. So, for 2020 years, the formula becomes:

A = $4000(1 + 0.06/1)^(1*2020)

Calculating this would give us the amount of money in the bank at the end of 2020 years.

This problem has been solved

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