2.4 Exponential and Logarithmic Functions
Key Takeaways
- Exponential growth P(t) = a·b^(t/k) counts how many periods of length k fit in t; doubling every 6 hours from 120 gives P(t) = 120(2)^(t/6).
- A logarithm log_b(x) = y is equivalent to b^y = x, requiring x > 0 and base b > 0 with b ≠ 1.
- Solving log₃(x - 1) = 2 yields x - 1 = 9, so x = 10, which satisfies the domain restriction x > 1.
- Exponential and logarithmic functions with the same base are inverse reflections across the line y = x.
- Praxis 5165 frequently tests contextual interpretation of growth factor, initial value, and period length.
Why This Section Matters
Exponential and logarithmic functions model growth, decay, and inverse relationships across Praxis 5165. Expect equation solving, interpretation of bases and exponents, half-life and doubling contexts, and links between y = b^x and y = log_b(x). Logarithms undo exponentials — ETS tests that inverse relationship directly and in teaching scenarios about student errors with log properties.
Exponential Functions
Standard form: f(x) = a · b^x with b > 0, b ≠ 1.
| Base b | Behavior |
|---|---|
| b > 1 | Exponential growth |
| 0 < b < 1 | Exponential decay |
a is often the initial value when x = 0 in untransformed models.
Worked Example: Doubling Growth
A culture starts with 120 bacteria and doubles every 6 hours. Model P(t) after t hours.
Each 6-hour interval multiplies by 2, so the number of doubling periods is t/6:
P(t) = 120(2)^(t/6)
Verify: P(6) = 120(2) = 240; P(12) = 120(4) = 480.
Common wrong models:
- P(t) = 120(2)^(6t) — grows far too quickly
- P(t) = 120(2t)^6 — places the variable in the base incorrectly
- P(t) = 120 + t/6 — linear, not exponential
Percent Change Form
Discrete percent growth: A = P(1 + r)^t. If a population grows 8% per year from P = 500, then A(t) = 500(1.08)^t.
Continuous growth uses A = Pe^(rt) with e ≈ 2.718. Recognize e as the continuous-compounding limit even when items stay discrete.
Logarithmic Functions
y = log_b(x) means b^y = x, with x > 0, b > 0, b ≠ 1.
Key Properties
| Property | Rule |
|---|---|
| Inverse | log_b(b^x) = x and b^(log_b x) = x |
| Product | log_b(xy) = log_b x + log_b y |
| Quotient | log_b(x/y) = log_b x - log_b y |
| Power | log_b(x^n) = n log_b x |
Change of base: log_b x = (ln x)/(ln b)
Worked Example: Solving a Log Equation
Solve log₃(x - 1) = 2.
Rewrite exponentially: x - 1 = 3² = 9 → x = 10.
Domain check: argument x - 1 > 0 requires x > 1. Since 10 > 1, the solution is valid.
Trap: 9 is x - 1, not x.
Graph Relationships
y = 2^x and y = log₂(x) are reflections across y = x.
Anchor points:
- Exponential 2^x passes through (0, 1)
- Logarithm log₂(x) passes through (1, 0)
Domain of log_b x is x > 0; range is all reals. Domain of b^x is all reals; range is y > 0.
Decay and Half-Life
P(t) = P₀(0.5)^(t/H) halves every H units. In P(t) = 120(0.5)^(t/8), the half-life is 8 hours, initial value 120, decay factor ½ per period.
Identify three pieces on context items:
- Initial value
- Growth or decay factor per period
- Period length in the exponent denominator
Teaching Connections
Students often write log(x + y) = log x + log y. The correct product rule applies to log(xy), not addition inside the argument. A quick counterexample: log(2 + 3) ≠ log 2 + log 3.
Common Praxis Traps
- Putting t in the base instead of the exponent for doubling problems
- Forgetting x > 0 (or argument > 0) for logarithms
- Confusing log₃(x - 1) = 2 solution 10 with 9
- Treating ln e^x as x ln e without simplifying (ln e = 1)
Section Takeaways
Build exponentials by counting growth periods in the exponent; solve logs by rewriting in exponential form with domain checks. Practice contextual modeling on /practice/praxis-math.
Solving Exponential Equations
To solve 2^(x+1) = 32, rewrite with a common base: 32 = 2⁵, so x + 1 = 5 and x = 4.
When bases cannot match, take logarithms of both sides:
3^x = 10 → x log 3 = log 10 → x = (log 10)/(log 3).
Praxis 5165 may present the answer in exact log form rather than a decimal.
Worked Example: Half-Life
A substance has half-life 5 years. Initial mass 80 g. Model remaining mass M(t) after t years:
M(t) = 80(0.5)^(t/5)
After 10 years: M(10) = 80(0.5)² = 20 g.
Logarithmic Scales (Conceptual)
Log scales compress large ranges — pH, Richter, and decibel scales are logarithmic. A Praxis item might ask which statement implies multiplicative change rather than additive; link that to exponential/log structure.
Expanding and Condensing Logs
Expand log(5x²/√y):
log 5 + log(x²) - log(√y) = log 5 + 2 log x - ½ log y
Condensing reverses the process. These algebra moves support solving equations and interpreting transformed data.
Inverse Function Table
| Exponential | Logarithmic |
|---|---|
| y = 2^x | y = log₂(x) |
| Domain: all reals | Domain: x > 0 |
| Range: y > 0 | Range: all reals |
| Passes (0, 1) | Passes (1, 0) |
Swapping x and y in an exponential equation produces the log form — a quick check when students confuse bases.
Compound Interest Connection
If money earns 6% annual interest compounded once per year, A = 1000(1.06)^t after t years from principal 1000. Compounding n times per year uses A = P(1 + r/n)^(nt). As n → ∞, the model approaches Pe^(rt) — the continuous case.
Praxis items may ask you to identify which parameter controls the growth rate (r or base b) versus the starting amount (P or a).
Checking Solutions
Always substitute back:
For log₃(x - 1) = 2, x = 10 gives log₃(9) = 2 because 3² = 9 ✓.
For 2^x = -4, no real solution exists because 2^x > 0 for all real x. Recognizing impossible exponential equations prevents wasted algebra on test day.
A culture starts with 120 bacteria and doubles every 6 hours. Which function models the population after t hours?
Solve log₃(x - 1) = 2.