Iva deposits $2,000 into an interest-bearing savings account that is compounded continuously at an interest rate of 5%. She decides not to deposit or withdraw any money after the initial deposit. We can represent the account balance of the savings account after t years by an exponential function:

A (t) = $2,000 ∙ e^0.05t.

Approximately how many years will it take for the initial deposit to double?

Question

Mathematics

Buzz

6 May, 19:45

Iva deposits $2,000 into an interest-bearing savings account that is compounded continuously at an interest rate of 5%. She decides not to deposit or withdraw any money after the initial deposit. We can represent the account balance of the savings account after t years by an exponential function:

A (t) = $2,000 ∙ e^0.05t.

Approximately how many years will it take for the initial deposit to double?

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Harper Shaw · Answer 1

13.863 years

Step-by-step explanation:

Initial deposit is $2,000.

The rate of interest is 5% compounded continuously.

The account balance of the savings account after t years by an exponential function:

A (t) = $2,000 ∙ e^0.05t.

It says to find out the time when it takes for the initial deposit to double, i. e. A (t) = $4,000

Mathematically, we can set them equal and solve for t as follows:-

A (t) = $2,000 ∙ e^0.05t = $4,000.

e^ (0.05t) = 4000/2000 = 2

0.05t Ln (e) = Ln (2)

t/20 = Ln (2)

t = 20 * Ln (2) = 13.86294361

So, it takes 13.863 years for the initial deposit to double.