
Bradford T. answered 11/08/21
Retired Engineer / Upper level math instructor
a)
A=A0ekt
After 10 years
45000 = 30000e10k
k= ln(1.5)/10
A(t) = 30000(1.5)k/10
In 2023, after 15 years
A(15) = 30000(1.5)1.5 = £55113.52
b)
A=A0(1-r)t
A(t) = 50000(1-.01)t = 50000(0.99)t
A(12) = 50000(0.99)12 = £44319.24
Change = (50000-44319.24)/50000 = 0.1136 = 11.36%