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 min TEKS 5A,5B Algebra 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.

The natural base e

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

Continuous compounding

A = Pert P = principal, r = rate (decimal), t = time For 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.

Practice

Continuous compounding formula

Which represents continuous compound interest of $P at rate r for t years?

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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/h A0 = initial amount, h = half-life period, t = elapsed time

Worked example

Half-life 10 years, elapsed 40 years Number of half-lives = 40 / 10 = 4 Fraction remaining = (1/2)4 = 1/16
Practice

Half-life over 4 periods

A radioactive isotope has a half-life of 10 years. What fraction remains after 40 years?

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Solving exponential equations (without logs)

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

3x = 81 3x = 34    (rewrite 81 as 34) x = 4 When the bases match, the exponents must match. (For different bases, you need logarithms — next lesson.)
Practice

Same-base trick

Solve: 3ˣ = 81.

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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).