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Problem:



Solution:

a)
$ f(t) = 10000 \times (1 +\frac{0.0242}{12})^{12t} $

b)
$ f(20) = 16217.61042 $

c)
Annual growth factor = $ (1 +\frac{0.0242}{12})^{12} = 1.024470231 $
Annual percentage change = $ 1.024470231 - 1 = 0.024470231 = 2.447023091\% $

d)
We want to find out the $ t $ such that $ f(t) = 20000 $, solving using logarithms, get $ t = 28.67131691 $

e)
That would become $ 10000 \times (1 +\frac{0.0242}{365})^{365 \times 20} = 16225.25613 $