AP PRECALCULUS — UNIT 2B · EXPONENTIAL & LOGARITHMIC FUNCTIONS
2.13B — Exponential and Logarithmic Equations and Inequalities
Notes — Solving with Logarithms and Analyzing Inequalities
💡 Learning Objectives (2.13.B)
By the end of this lesson you will be able to:
- Solve exponential equations whose bases cannot be matched, by taking a logarithm of both sides
- Solve more complex logarithmic equations and check for extraneous solutions
- Solve exponential and logarithmic inequalities and express solutions in interval notation
- Reason about the validity of solutions in modeling contexts
1. When the Bases Cannot Be Matched
If you have something like 5^x = 17, no clever rewriting puts both sides in the same base. The strategy: take a logarithm of both sides and use the power rule to bring the exponent down.
📘 Example — Solve 5^x = 17
- Take ln of both sides: ln(5^x) = ln(17)
- Apply the power rule: x · ln(5) = ln(17)
- Solve: x = ln(17) / ln(5) ≈ 1.7604
- Check: 5^1.7604 ≈ 17 ✓
📝 Which logarithm to use?
It does not matter — log, ln, or any base produces the same answer. Use ln when the base is e, log when the base is 10, and either one in all other cases. Both give the same numerical value via change of base.
2. Equations with the Variable on Both Sides
📘 Example — Solve 3^(2x) = 7^(x+1)
- Take ln of both sides: 2x · ln(3) = (x + 1) · ln(7)
- Distribute: 2x · ln(3) = x · ln(7) + ln(7)
- Group x-terms: 2x · ln(3) − x · ln(7) = ln(7)
- Factor: x · (2 ln(3) − ln(7)) = ln(7)
- Solve: x = ln(7) / (2 ln(3) − ln(7)) ≈ 7.61
3. Equations Quadratic in Form
Some exponential equations look quadratic once you substitute u = b^x:
📘 Example — Solve e^(2x) − 5e^x + 6 = 0
- Let u = e^x. Then u² − 5u + 6 = 0
- Factor: (u − 2)(u − 3) = 0, so u = 2 or u = 3
- Back-substitute: e^x = 2 ⇒ x = ln(2); e^x = 3 ⇒ x = ln(3)
- Both are valid (e^x is always positive).
4. More Complex Logarithmic Equations
📘 Example — Solve ln(x + 5) − ln(x − 1) = ln(3)
- Condense the left: ln((x + 5)/(x − 1)) = ln(3)
- Same base both sides; set arguments equal: (x + 5)/(x − 1) = 3
- Solve: x + 5 = 3(x − 1) = 3x − 3, so 2x = 8 and x = 4
- Check: x − 1 = 3 > 0 ✓ and x + 5 = 9 > 0 ✓
5. Exponential Inequalities (Different Bases)
📘 Example — Solve 2^x > 100
- Take ln of both sides (preserves direction since ln is increasing): x · ln(2) > ln(100)
- Divide by ln(2) > 0 (no flip): x > ln(100)/ln(2) ≈ 6.6439
- Solution: x > log₂(100) ≈ 6.6439, or in interval notation (log₂(100), ∞)
⚠️ Common mistake
If you take a log with base < 1 (or multiply through by a negative quantity such as ln(0.5) ≈ −0.693), you must reverse the inequality. Stick to natural log or base-10 log to keep the multiplier positive.
6. Logarithmic Inequalities
Two simultaneous conditions matter: the inequality itself, and the requirement that the argument of every log be positive.
📘 Example — Solve log₃(x − 2) < 4
- Domain restriction: x − 2 > 0, so x > 2
- Rewrite: x − 2 < 3⁴ = 81 (base 3 > 1, direction preserved)
- So: x < 83
- Combine: 2 < x < 83, or (2, 83) in interval notation
7. Validity in Modeling Contexts
In an applied problem, time t is usually nonnegative, populations are positive integers, and so on. After solving algebraically, look at each candidate to see whether it makes physical sense. A negative time value or a fractional ‘number of bacteria’ may need to be rounded or rejected.
8. Summary
- Take a log of both sides to solve exponential equations whose bases differ
- Use the power rule to bring exponents down so you can solve linearly
- Substitute u = b^x to convert quadratic-in-form exponentials to a familiar quadratic
- Inequality direction depends on base: > 1 preserves, < 1 reverses
- Always intersect your algebraic solution with the domain requirements