interest rate???

Question

If you invest $1000 in an account that pays interest at a rate of 6.5% per year continuously compounded. How long would it take to double your money? See hint for the formula, where A is the total balance, r is the interest rate, and t is the number of years.

 

Hint:
a(t)=pe^rt

Answer 1

Hi Daniel,

 

Let p =principal

     A = Amount or total balance

     r= interest rate

     m = number of compoundings per period

     t = number of years

 

p= $1000

A= 2P = 2000

r = 6.5% = 0.065

m =∞

 

when you have a continuous compounding case, we use the following formula to compute the total balance, A.

 

               A = Pe^rt; where e is the base of natural logarithm, ln.

              2000 = 1000e^0.065t

divide both sides by 1000, we have

                2 = e^0.065t

Taking the natural logarithm of both sides, we have

              ln2 = ln(e^0.065t)

              ln2 = 0.065t(lne).....the natural logarithm of e is 1

           0.69315 = 0.065t (1)

                  t = 0.69315/.065

                    = 10.663 years

Therefore, it takes 10 years and 8 months for $1,000 to double itself in value if it is invested in an account that pays 6.5% compounded instantaneously.

 

Thanks,

A.

 

           

Answer 2

2000 = 1000e^.065t

 

2 = e^.065t

 

ln (2) = .065t

 

t = ln(2)/.065