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Learn · Algebra 2

Exponential Functions Advanced: Continuous Growth, e, and Half-Life

Algebra 1 introduced exponential growth. Algebra 2 deepens it: the natural base e, continuous compounding, and decay problems where the half-life is measured in years.

9 minTEKS 5A,5BAlgebra 2

Discrete to continuous

Algebra 1 used P(1 + r)t for compound interest applied annually. Algebra 2 introduces Pert for continuous compounding — what happens when interest is added every instant. The natural base e ≈ 2.71828 is the unique number that makes calculus on exponential functions clean.

Exponential: growth if b > 1, decay if 0 < b < 1H.A. y = 0 (both curves)y = 2ˣ (growth)y = (½)ˣ (decay)(0, 1)

The natural base e

e = limn→∞ (1 + 1/n)n ≈ 2.71828e arises naturally as the limit of compounding more and more often. The function f(x) = eˣ is its own derivative.

Continuous compounding

A = PertP = principal, r = rate (decimal), t = timeFor 5% annual rate over 3 years: A = P · e0.05·3 = P · e0.15 ≈ 1.162 P
Annual vs continuous

For the same rate and time, continuous compounding always yields slightly more than annual: er > (1 + r). The gap grows with the rate.

Continuous compounding formula
Which represents continuous compound interest of $P at rate r for t years?
Compound interest: A = P(1 + r/n)^(nt); continuous: A = Pe^(rt)Compound n times per yearA = P(1 + r/n)^(nt)P: principal, r: rate, n: times/yr, t: years$1000, 5%, monthly, 10y → $1647Continuous compoundingA = P · e^(rt)as n → ∞, formula → this$1000, 5%, cont., 10y → $1649

Half-life: decay problems

For radioactive isotopes (and many other decay processes), every fixed time period multiplies the remaining amount by ½.

A(t) = A0 · (1/2)t/hA0 = initial amount, h = half-life period, t = elapsed time

Worked example

Half-life 10 years, elapsed 40 yearsNumber of half-lives = 40 / 10 = 4Fraction remaining = (1/2)4 = 1/16
Half-life over 4 periods
A radioactive isotope has a half-life of 10 years. What fraction remains after 40 years?

Solving exponential equations (without logs)

If you can rewrite both sides with the same base, just equate the exponents.

3x = 813x = 34 (rewrite 81 as 34)x = 4When the bases match, the exponents must match. (For different bases, you need logarithms — next lesson.)
Same-base trick
Solve the exponential equation 3ˣ = 81 for x.
Half-life: A = A₀·(½)^(t/t½) — precisely exponential (amount halves every t½)100%1 t½50%2 t½25%3 t½12%4 t½6.25%%Each half-life reduces the amount by exactly ½; never reaches 0 (asymptotic decay)

3-second recap

  • e ≈ 2.71828 is the natural base. A = Pert for continuous compounding.
  • Half-life: number of half-lives = elapsed time ÷ half-life period; multiply by (1/2)that many.
  • Same base on both sides → equate exponents.
  • Different bases → use logarithms (next lesson).

Check yourself

Quick check #1
A radioactive substance has a half-life of 10 years. After 30 years, what fraction remains?
Quick check #2
Which formula models continuous compound interest?