Question

Suppose that $17,699 is invested at an interest rate of 6.6% 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?​

Answer 1

a)

$s(t) = 17699(1 .066) {}^{t}$

b)

$s(1) = 17699(1.066) = 18867.13 \\ s(2) = 17699(1.066) {}^{2} = 20112.36 \\ s(5) = 17699(1.066) {}^{5} = 24363.22 \\ s(10) = 17699(1.066) {}^{10} = 33536.73$

c)

$s(t) = 2 \times initial \: capital \: \\ s(t) = 2 \times 17699$

$17699(1.066) {}^{t} = 2 (17699) \\ 1.066 {}^{t} = 2 \\ t = log¹°⁰⁶⁶(2) = 10.84511 \: years$