f[0] = P
f[t+1] = f[t] + 0.12 f[t]
Now we may use the distributive rule to rewrite the second equation as
f[t+1] = 1.12 f[t]
and then use it to deduce:
f[1] = f[0+1] = 1.12 f[0] = 1.12 P
and
f[2] = f[1+1] = 1.12 f[1] = 1.12 1.12 P
and also
f[3] = f[2+1] = 1.12 f[2] = 1.12 1.12 1.12 P
and so on. The general pattern is recognizable - after t years we will have
multiplied P by 1.12 t times, and that is the same as multiplying by the t-th
power of 1.12, so we get this formula:
t
f[t] = 1.12 P
Now until now we were talking about values of t which are integers (whole
numbers), but if we let t take on arbitrary real values we can use the same
formula to predict values of f[t] at any point within a year.
Finally, to relate this to the natural logarithm as introduced in the text, we use the definition of Log[1.12] as that number which you have to raise E to to get 1.12, i.e., we have this formula:
Log[1.12]
1.12 = E
This, together with the high school algebra identity
a b (a b)
(x ) = x
allows us to write the following formula.
Log[1.12] t
f[t] = P E