As stated in the text, the "Rule of 70" says that inflation will double the price level in (70/rate of inflation) years. For example, annual inflation of 5% will cause prices to double in (70/5) = 14 years. More generally, any initial amount, whether it be the price level or the value of your Individual Retirement Account (IRA), will double in value in 70/(gx100%) years, where g is its annual growth rate. Why does this formula work?

Suppose we start with some initial amount, say V₀. Growing at a rate of gx100%, it will have added gV₀ to its value at the end of the first year. If V₁ is the value at the end of the first year, then V₁ = V₀ + gV₀ = V₀(1 + g). Using similar reasoning, its value at the end of the second year will be V₂ = V₁(1 + g). But since V₁ = V₀(1 + g), we can substitute for V₁ to obtain V₂ = V₀(1 + g)². By induction, we can see that after T years, the initial value will have grown to V_T = V₀(1 + g)^T.

The question at hand is this: at what value of T will V_T = 2V₀? Substituting 2V₀ for V_T, 2V₀ = V₀(1 + g)^T, or upon dividing by V₀, 2 = (1 + g)^T. Our task now is to solve this for T in terms of g. If we take the natural logarithm of both sides, we get ln(2) = Tln(1 + g). Next we make use of an important result: for very small values of g, as we might find for reasonable rates of inflation or returns on investments, ln(1 + g) is approximately equal to g. (You may recall that ln(1) = 0.) Making this substitution, and noting that ln(2) = 0.693147... » .70, we have our result: T » .70/g. If g is expressed as a percentage, simply multiply both top and bottom by 100: T » 70/(gx100%).

Incidentally, the rough approximation inherent in ln(1 + g) » g disappears when there is continuous exponential compounding of the growth, rather than annual compounding. With continuous compounding, it can be shown that after T years, an initial value of V₀ will grow to V_T = V₀e^gT where e is the base of natural logarithms. As before, we wish to know the value of T for which V_T = 2V₀, or equivalently, solve for T in the equation 2 = e^gT. Taking the natural logarithm of both sides, we obtain ln(2) = gT, or T = 0.693147.../g » 70/(gx100%) as before.