Question

You deposit $6000 in an account earning 4% interest compounded continuously.

The amount of money, A, in the account after t years is given by A=6000e^0.04T.

How much will you have in the account in 15 years?
I got 10932.71 with rounding to 2 decimal places but think i am wrong

How long will it be until you have $18500 in the account? Round to 2 decimal places.
_____ years?

How long does it take for the money in the account to double? Round to 2 decimal places
_____ years?

Answer 1

Explanation:

For the first question, you need to calculate the amount of money in the account after 15 years. This is done by plugging 15 into the equation for A.

A=6000e^0.04T

A = 6000e^0.6

A = $10,098.45

For the second question, you need to calculate how long it takes until the account has $18500. To do this, you need to solve the equation for T.

A=6000e^0.04T

18500 = 6000e^0.04T

18500/6000 = e^0.04T

ln(18500/6000) = 0.04T

T = ln(18500/6000)/0.04

T = 24.64 years

For the last question, you need to calculate how long it takes for the money in the account to double. This can be done by setting A equal to 12000 and solving for T.

A=6000e^0.04T

12000 = 6000e^0.04T

12000/6000 = e^0.04T

ln(12000/6000) = 0.04T

T = ln(12000/6000)/0.04

T = 17.31 years

Answer 2

a.) 10,098.45

b.)24.64 years

c.)17.31 years

Explanation:

For the first question, you need to calculate the amount of money in the account after 15 years. This is done by plugging 15 into the equation for A.

A=6000e^0.04T

A = 6000e^0.6

A = $10,098.45

For the second question, you need to calculate how long it takes until the account has $18500. To do this, you need to solve the equation for T.

A=6000e^0.04T

18500 = 6000e^0.04T

18500/6000 = e^0.04T

ln(18500/6000) = 0.04T

T = ln(18500/6000)/0.04

T = 24.64 years

For the last question, you need to calculate how long it takes for the money in the account to double. This can be done by setting A equal to 12000 and solving for T.

A=6000e^0.04T

12000 = 6000e^0.04T

12000/6000 = e^0.04T

ln(12000/6000) = 0.04T

T = ln(12000/6000)/0.04

T = 17.31 years