🌳 Population Growth & Depreciation

Growth (appreciation): a population or value rising at R% per year for n years becomes $P\left(1+\dfrac{R}{100}\right)^n$. You simply multiply by the growth factor once for each year — a 10% rise means $\times 1.1$, a 5% rise means $\times\dfrac{21}{20}$.

Decay (depreciation): cars, machines and assets lose value, so they follow $P\left(1-\dfrac{R}{100}\right)^n$. Each year you multiply by what is left: a 10% drop means $\times 0.9$, a 20% drop means $\times\dfrac{4}{5}$, a 25% drop means $\times\dfrac{3}{4}$.

Running time backwards: to find a past population or original price, do the opposite — divide the present value by the factor instead of multiplying. Never just “subtract the loss back”, because each year's change sits on a different base.

Different rates in different years? Just multiply each year's own factor in turn. A value that falls 10% one year and 15% the next becomes $P\times 0.9\times 0.85$ — handle each year separately, then combine.

The trap: adding a flat $R\times n$ % (e.g. “20% off for two 10% years”) is always wrong. For growth it undercounts, for decay it overcounts — the base keeps changing every year.