Question 1120705: Suppose that $10,095 is invested at an interest rate of 6.2% per year, compounded continuously.
a) Find the exponential function that describes the amount in the account after time t, in years.
b) What is the balance after 1 year? 2 years? 5 years? 10 years?
c) What is the doubling time?
I don't know how to set up the formula. Once the formula is set up I know that I substitute the number into t.
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Suppose that $10,095 is invested at an interest rate of 6.2% per year, compounded continuously.
a) Find the exponential function that describes the amount in the account after time t, in years.
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V(t) = P*e^t
P = 10,095, e = 2.718281828459045....
b) What is the balance after 1 year? 2 years? 5 years? 10 years?
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c) What is the doubling time?
Solve e^t = 2
t = ln(2)/ln(e) = ln(2) = 0.693 yrs
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I don't know how to set up the formula. Once the formula is set up I know that I substitute the number into t.
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Cheers,
Stan H.
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Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! Suppose that $10,095 is invested at an interest rate of 6.2% per year, compounded continuously.
a) Find the exponential function that describes the amount in the account after time t, in years.
b) What is the balance after 1 year? 2 years? 5 years? 10 years?
c) What is the doubling time?
I don't know how to set up the formula. Once the formula is set up I know that I substitute the number into t.
Correct formula:  , where:
P = $10,095
e = 2.7182818.......
r = Annual interest rate (6.2%, or .062, in this case)
t = time
What is wrong with these people on here? How can an amount take less than 1 YEAR to double? RIDICULOUS!!
At a rate of 6.2%, and with CONTINUOUS compounding, ANY investment/deposit will take  to DOUBLE!
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