SOLUTION: Suppose that $2000 is invested at a rate of 6% per year compounded continuously. What is the balance after 1 yr? After 2 yrs?

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Question 332248: Suppose that $2000 is invested at a rate of 6% per year compounded continuously. What is the balance after 1 yr? After 2 yrs?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Recall that the formula for continuous compounding interest is

A=Pe%5E%28rt%29

where A is the return, P is the principal (amount invested), r is the interest rate (in decimal form) and t is the time in years.


Since "$2000 is invested at a rate of 6% per year compounded continuously", we know that P=2000 and r=0.06 (the decimal equivalent of 6%).


Now let's compute the return when t=1 (ie find the balance after one year)


A=Pe%5E%28rt%29 Start with the continuous compounding formula.


A=2000%2Ae%5E%280.06%2A1%29 Plug in P=2000, r=0.06, and t=1.


A=2000%2Ae%5E%280.06%29 Multiply 0.06 and 1 to get 0.06.


A=2000%2A1.06183654654536 Raise 'e' (which is approximately 2.71828) to the power 0.06 to get 1.06183654654536 (this value is approximate).


A=2123.67309309072 Multiply 2000 and 1.06183654654536 to get 2123.67309309072.


A=2123.67 Round to the nearest hundredth (ie to the nearest penny).


So after 1 year, you'll have about $2,123.67


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Now let's compute the return when t=2 (ie find the balance after two years)


A=Pe%5E%28rt%29 Start with the continuous compounding formula.


A=2000%2Ae%5E%280.06%2A2%29 Plug in P=2000, r=0.06 (the decimal equivalent of 6%), and t=2.


A=2000%2Ae%5E%280.12%29 Multiply 0.06 and 2 to get 0.12.


A=2000%2A1.12749685157938 Raise 'e' (which is approximately 2.71828) to the power 0.12 to get 1.12749685157938 (this value is approximate).


A=2254.99370315875 Multiply 2000 and 1.12749685157938 to get 2254.99370315875.


A=2254.99 Round to the nearest hundredth (ie to the nearest penny).


So after two years, you'll have $2,254.99