f=ir^t, f=final amount, i=initial amount, r=common ratio, t=time
We are told that r=1.1 and i=100 so
f=100(1.1^t) so when t=2.25
f=100(1.1^2.25)
f≈$123.92 (to nearest cent)
You may also see someone using:
A=Pe^(kt) Since the yearly rate is 10% you know that after 1 year there will be 110 so we can solve for k
110=100e^(k)
1.1=e^k if we take the natural log of both sides
ln(1.1)=k, and again P=100 so
A=100e^(t*ln1.1), so when t=2.25
A=100e^(2.25*ln1.1)
A≈$123.92
The problem with the so-called continuous compound formula is that it is no more accurate than the neater f=ir^t. And because of the k value which is really a natural log cannot be expressed as a rational number in most cases, and is then "rounded" leaving errors in the approximation. k values aren't as neat as r values. The advantage of A=Pe^(kt) comes in handy when differential equations are used, but other than that, there is no advantage when simply calculating exponential growth/decay.