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Suppose that​\$10,055 is invested at an interest rate of 5.6​% per​ year, compounded continuously.

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Suppose that​\$10,055 is invested at an interest rate of 5.6​% 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?

Guest Feb 26, 2017
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\(A = Pe^{rt}\)

e is about equal to 2.71828182846...

a) \(A = 10,055e^{0.056*year}\)

b)

1 Year = \$10634.14

2 Years = \$11246.64

5 Years = \$13304.07

10 Years = \$17603.01

c) \(10055*2=10055e^{0.056t}\)

\(2=e^{0.056t}\)divide both sides by 10055

\(ln 2 = 0.056t\)take the natural log of both sides

\(ln2/0.056 = t\)divide both sides by 0.056

\(ln 2 = log_e 2\)ln is just natural log, which is log base e

t = 12.3776282242847375

Guest Feb 27, 2017

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