More investments

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