Suppose you deposit $1,000 in a savings account that pays interest at an annual rate of 4%. If no money is added or withdrawn from the account, answer the following questions.
a. After 3 years, the amount in the account will be how much (Do not round until the final answer. Then round to the nearest cent as needed)
b. After 17 years, the amount in the account will be how much (Do not round until the final answer. Then round to the nearest cent as needed)
c. For the account to contain $1,500, it will take how many years. (Type a whole number).
d. For the account to contain $2,000 it will take how many years. (Type a whole number).
a. To calculate the amount in the account after 3 years, we can use the formula for compound interest: A = P(1 + r/n)^(nt), where A is the final amount, P is the principal amount (initial deposit), r is the annual interest rate (as a decimal), n is the number of times the interest is compounded per year, and t is the number of years.
Plugging in the given values: A = 1000(1 + 0.04/1)^(1*3) = 1000(1.04)^3
Calculating the expression: A ≈ 1000(1.124864) ≈ $1,124.86
b. Using the same formula: A = 1000(1 + 0.04/1)^(1*17) = 1000(1.04)^17
Calculating the expression: A ≈ 1000(1.848582) ≈ $1,848.58
c. Let's set up the equation to find the number of years when the account contains $1,500: 1,000(1 + 0.04/1)^(1*x) = 1,500
Simplifying the equation: (1.04)^x = 1.5
Taking the logarithm of both sides: x*log(1.04) = log(1.5)
Solving for x: x = log(1.5)/log(1.04)
Calculating the expression: x ≈ 23.26
Therefore, it will take approximately 23 whole years for the account to contain $1,500.
d. Following the same steps as in part c, we set up the equation: 1,000(1 + 0.04/1)^(1*x) = 2,000
Simplifying the equation: (1.04)^x = 2
Taking the logarithm of both sides: x*log(1.04) = log(2)
Solving for x: x = log(2)/log(1.04)
Calculating the expression: x ≈ 34.66
Therefore, it will take approximately 34 whole years for the account to contain $2,000.