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Exponential and Logarithmic Functions: Inverses That Run Your World

Why exp and log are inverses, the three log rules that solve everything, and the domain trap that sinks log equations.

8 minTEKS 3E-3IPre-Calculus

Exponential and logarithm are inverses

If b > 0 and b ≠ 1:

bˣ = y ⟺ log_b(y) = x

So e^(ln a) = a and ln(eᵇ) = b. They undo each other.

y = 2ˣ and y = log₂(x): inverse functions, reflected across the line y = x12−112−1y = xy = 2ˣy = log₂(x)(0, 1)(1, 0)← log asymptote at x = 0exp: domain ℝ, range (0,∞), y-intercept (0, 1) · log: domain (0,∞), range ℝ, x-intercept (1, 0)Because they are inverse functions, (a, b) on one ↔ (b, a) on the other

The three log rules

  • Product: log_b(MN) = log_b M + log_b N
  • Quotient: log_b(M/N) = log_b M − log_b N
  • Power: log_b(Mⁿ) = n · log_b M

And the identities: log_b(b) = 1, log_b(1) = 0, log_b(bᵏ) = k.

Solving exponential equations

If you can match bases, equate exponents: 2x+1 = 32 = 2⁵ → x + 1 = 5 → x = 4.

If you can't, take log of both sides: 3x = 7 → x = ln(7)/ln(3) ≈ 1.77.

Log properties: multiplication becomes addition, powers become multipliersProduct rulelog(xy) = log x + log ymultiply inside → add outsideQuotient rulelog(x/y) = log x − log ydivide inside → subtract outsidePower rulelog(xⁿ) = n · log xexponent → coefficient

The log-equation domain trap

⚠️ Always check domain

After solving ln(x) + ln(x − 3) = ln(10) → x = 5 or −2, the −2 is EXTRANEOUS because ln(−2) is undefined. Always check that the original log arguments are positive.

Growth and decay models

Continuous compounding / population growth: A(t) = A₀ · ert. For r > 0 grows; r < 0 decays. Half-life formula: t½ = ln(2)/k.

Natural base e ≈ 2.71828 — the limit of continuous compoundingy = e ≈ 2.718(1 + 1/n)ⁿn=1: (1+1)¹ = 2n=12: 2.613n=∞: e = 2.71828…

Check yourself

📌 Solve
If 4x = 64, what is x?

Practice with CBE-style practice questions

Try exp/log questions in Pre-Calc Sem A practice.