Find the amount of money that results if $100 is invested at 10% compounded continuously for 2.25 years.

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.  

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