Why small rates become big differences

Sep 2026 · 3 min read · math, economics

Sample post: mock editorial content for this portfolio, not a record of personal research.

Five percent and seven percent sound close. Over one year, they are. Over several decades, the difference becomes much harder to ignore.

The reason is not that the arithmetic gets complicated. It is that each year's growth changes the starting point for the next.

Growth builds on growth

Start with an amount $P>0$ and a constant annual growth rate $r$ . After one year, the amount is $P(1+r)$ . After two, it is $P(1+r)^2$ . After $n$ whole years:

A_n = P(1+r)^n.

This model assumes annual compounding, no deposits or withdrawals, and no fees or taxes. It is a mathematical illustration, not a forecast of investment returns.

For an initial 1,000 units growing at 5% a year for 20 years:

A_{20} = 1000(1.05)^{20} \approx 2653.30.

At 7%, the same starting amount becomes approximately 3,869.68 units. That is about 46% more than the 5% outcome, even though the annual rates differ by only two percentage points.

One useful way to see the difference is to cancel the starting amount:

\frac{A_{20}^{(7\%)}}{A_{20}^{(5\%)}} = \left(\frac{1.07}{1.05}\right)^{20} \approx 1.4584.

The relative advantage comes from the rates and the time horizon, not the size of the initial balance.

Logarithms turn time into something solvable

Suppose we want to know how long doubling takes. For a positive rate $r$ , extend the growth formula to a real-valued time $t$ and set the final amount equal to twice the starting amount:

2P=P(1+r)^t.

Cancel $P$ and take natural logarithms:

t=\frac{\ln 2}{\ln(1+r)}.

At 5%, this is about 14.21 years; at 7%, about 10.24 years. If interest is credited only at whole-year intervals, the first doubled balances occur at years 15 and 11, respectively. The fractional answers describe the smooth interpolation between those annual observations.

For small positive rates, $\ln(1+r)\approx r$ , so:

t\approx\frac{0.6931}{r}.

Here $r$ is a decimal, not a percentage number: use $0.05$ , not $5$ . The approximation gives 13.86 years at 5%, slightly below the exact answer. It is a shortcut, not an identity.

Continuous growth is the same idea at a different scale

If growth occurs continuously at a constant rate $g$ , the model becomes:

A(t)=Pe^{gt}.

To match an effective annual growth rate $r$ , choose $g=\ln(1+r)$ . A continuous rate of 5% is therefore not exactly the same as an effective annual rate of 5%.

This is why logarithms keep appearing in economics. They turn multiplicative changes into additive ones:

\ln\left(\frac{A_n}{P}\right)=n\ln(1+r).

The useful limit of the model

Real growth rates change. With annual rates $r_k>-1$ , the appropriate expression is:

A_n=P\prod_{k=1}^{n}(1+r_k).

A 10% gain followed by a 10% loss leaves $1.1\times0.9=0.99$ times the starting amount, not the original balance. Averaging the two percentages would hide that loss.

The takeaway is modest but useful: when changes build on one another, think in products. When you want to understand those products, reach for logarithms.