Exponentials: equal factors instead of equal increments

A hypothetical quantity starts at 100 and doubles every three hours: 200 at hour 3, 400 at hour 6. Adding 100 every three hours would give only 300. Exponential growth preserves a factor over equal times. Use power functions and progressions.

$E(t)=100\cdot2^{t/3},\quad L(t)=100+\frac{100}{3}t.\qquad E(t+3)=2E(t).$

At 1.5 hours E gives $100\sqrt2\approx141.42$ , not 150. Integer-period data alone do not specify intermediate values; this model additionally assumes consistent factors for fractions of periods.

Factor and increment

E(t)=100·2ᵗᐟ³; L(t)=100+100t/3

Solid E doubles per 3 h; dashed L adds 100 per 3 h. Shared initial data do not imply equal future growth. Hours, hypothetical quantity, uniform scales.

Real exponents

Fractional powers use roots. Irrational powers use the consistent continuous extension from rational approximations.

$a^{p/q}=\sqrt[q]{a^p}\ (a>0);\quad a^{u+v}=a^ua^v.$

$f(x)=a^x$ has domain ℝ and range (0,∞). It increases for a>1 and decreases for 0<a<1; a=1 gives constant 1. A positive base is required for arbitrary real exponents. OpenStax

Why e appears

Split a hypothetical 100% increase per time unit into m equal compounding steps. For m=1,2,10,100, $(1+1/m)^m$ is approximately 2; 2.25; 2.594; 2.705. It approaches e.

$\left(1+\frac1m\right)^m\longrightarrow e\approx2.71828;\qquad N(t)=N_0e^{kt}.$

Here k is a constant rate per time unit, and kt is dimensionless. Positive k gives growth, negative k decay. Doubling time satisfies $e^{kT}=2$ ; in our model T=3 hours. This does not prove unlimited real growth: resources and rates can change. Distinguish a model’s constant factor from a measured pattern.