Everything else in the chapter is assembled from this one relation. A present sum $P$ left for $n$ periods at rate $i$ becomes $F = P(1+i)^{n}$ — read aloud F equals P times one plus i, to the n. $P$ is the sum today in dollars, $F$ the sum at the end, $i$ the rate per period as a decimal, and $n$ the count of periods. You solve for $F$ going forward, for $P$ coming back, or for $n$ when both sums are known and the question is how long.

Read the shape and the whole subject falls out. $(1+i)$ is what one dollar becomes after one period. Do it $n$ times and you raise it to the $n$. The interest earned in period two is charged on the interest from period one — that is the entire difference between compound and simple interest, $F = P(1 + i\,n)$, which pays interest only on the original sum. Simple interest is a real thing on some short-term instruments and it is not what a reserve fund does. Over ten years at 8% the two differ by about a fifth of the balance, and the gap only widens.

Turned around, the same relation is discounting: $P = \dfrac{F}{(1+i)^{n}}$, the present worth of a future amount. Say plainly what that means, because it is misread constantly: it is not a prediction that the cost will shrink, and it is not inflation. It is the sum you would have to set aside today, at rate $i$, to have exactly $F$ waiting when the bill arrives. That is why a $150,000 media change in year 7 is not a $150,000 problem this year.

And when the unknown sits in the exponent, take logarithms: $n = \dfrac{\ln(F/P)}{\ln(1+i)}$. Note the shape — a ratio of two logs, never a product. If a stray minus appears, the fraction went in upside down.

Units doctrine, applied to money: the units here guide but never confess. Dollars in, dollars out, and both a compounded and a discounted answer wear dollars quite happily. Only the SIZE tells you which way you went — forward is always bigger, backward always smaller. Check the direction every single time.