Difficulty: Intermediate | Prerequisites: Functions and Polynomials study notes, basic exponent rules
Logarithms and exponentials are inverse operations, and most of what you do with them at this level involves moving between the two forms. This material sits between the functions unit and trigonometry: you need it for solving equations, simplifying compositions, and understanding growth models. If you are comfortable with exponent rules and the concept of an inverse function, you are ready.
Logarithm properties let you combine or break apart log expressions. Exponential equations are solved by isolating the exponential term and taking a logarithm. Logarithmic equations are solved by combining logs into one and exponentiating. Always check for extraneous solutions when logarithms are involved, because the argument of a log must be positive.
Natural logarithm (ln)
The logarithm with base e ≈ 2.718. ln x means "the power you raise e to in order to get x." In simple terms, ln and eˣ undo each other.
Exponential function
A function of the form f(x) = a · eᵏˣ or f(x) = a · bˣ. The variable is in the exponent, which is what makes these grow (or decay) so rapidly.
Logarithm properties
The three core rules for manipulating logarithms: the product rule (ln(ab) = ln a + ln b), the quotient rule (ln(a/b) = ln a − ln b), and the power rule (ln(aⁿ) = n ln a). Think of them as translating multiplication into addition, division into subtraction, and exponents into coefficients.
Extraneous solution
A value that emerges from the algebra but does not satisfy the original equation. In logarithmic equations, this happens when a candidate solution makes the argument of a logarithm zero or negative. You must always check.
Logistic equation
An equation of the form L / (1 + ae^(−kx)) = C, which models bounded growth. Solving it involves isolating the exponential term step by step.
The three core properties let you collapse multiple logarithm terms into one.
Product rule: ln a + ln b = ln(ab)
Quotient rule: ln a − ln b = ln(a/b)
Power rule: n · ln a = ln(aⁿ)
Example: Simplify ln(xy) + 2 ln(yz²) − ln(xz), assuming all variables are positive.
Apply the power rule first:
2 ln(yz²) = ln((yz²)²) = ln(y²z⁴)
Now combine using the product and quotient rules:
ln(xy) + ln(y²z⁴) − ln(xz) = ln(xy · y²z⁴) − ln(xz) = ln(xy³z⁴) − ln(xz)
= ln(xy³z⁴ / xz) = ln(y³z³)
The order of operations matters: apply the power rule before the product/quotient rules, since the power rule changes coefficients into exponents.
The strategy is to isolate the exponential term, then take the natural logarithm of both sides.
Example: Solve 44 / (1 + 4e^(−x/7)) = 32.
Step 1: Flip both sides (or cross-multiply) to isolate the exponential:
(1 + 4e^(−x/7)) / 44 = 1/32
1 + 4e^(−x/7) = 44/32 = 11/8
Step 2: Isolate the exponential:
4e^(−x/7) = 11/8 − 1 = 3/8
e^(−x/7) = 3/32
Step 3: Take ln of both sides:
−x/7 = ln(3/32)
x = −7 ln(3/32) = 7 ln(32/3)
The last step uses the property that −ln(a/b) = ln(b/a). Either form is correct, but the positive version is typically cleaner.
For logistic-type equations like this one, the algebra always follows the same pattern: reciprocate, isolate, logarithm.
The strategy: combine all logarithms into a single log, then exponentiate to remove it. After solving, check every candidate solution against the original equation's domain.
Example: Solve ln x − (1/2) ln(x + 4) = 0.
Step 1: Use log properties to combine:
ln x − ln((x + 4)^(1/2)) = 0
ln(x / √(x + 4)) = 0
Step 2: Exponentiate both sides:
x / √(x + 4) = e⁰ = 1
Step 3: Solve algebraically:
x = √(x + 4)
x² = x + 4 (squaring both sides)
x² − x − 4 = 0
x = (1 ± √17) / 2
Step 4: Check for extraneous solutions. The original equation requires x > 0 (for ln x) and x + 4 > 0 (always satisfied when x > 0).
x = (1 − √17)/2 ≈ −1.56, which is negative. This is extraneous.
x = (1 + √17)/2 ≈ 2.56, which is positive. This is the only valid solution.
Squaring both sides is the step that introduces extraneous solutions. Any time you square an equation, you must verify the results.
When eˣ and ln x are composed, they cancel. The key identities:
e^(ln u) = u (for u > 0)
ln(eᵘ) = u (for all u)
Example: f(x) = (1/2) ln(x + 2), g(x) = eˣ. Find (g ∘ f)(x).
(g ∘ f)(x) = g(f(x)) = g((1/2) ln(x + 2))
= e^((1/2) ln(x+2))
= e^(ln(√(x+2)))
= √(x + 2)
The exponent (1/2) ln(x + 2) simplifies to ln(√(x + 2)) by the power rule, and then e^(ln(…)) strips away the logarithm. The final answer should be free of both exponentials and logarithms.
Students often think ln(a + b) = ln a + ln b. It does not. The product rule says ln(a · b) = ln a + ln b. There is no "sum rule" for logarithms.
When solving logarithmic equations, students sometimes forget to check for extraneous solutions. Any time you square both sides or manipulate the equation in a way that could introduce false roots, you must substitute back into the original.
Students confuse ln(x²) with (ln x)². The first uses the power rule and equals 2 ln|x|. The second is a square of the logarithm. These are different expressions.
In logistic-type problems, students sometimes take the logarithm too early, before isolating the exponential term. The exponential must be alone on one side before you apply ln.
⚠️ Combining logarithms into a single expression is a near-certain exam question. Know all three properties cold.
⚠️ Solving exponential equations (particularly logistic-type) tests methodical algebra. Show every step.
⚠️ Extraneous solution checks are worth marks. If a question says "be sure to check for extraneous solutions," the examiner expects at least one candidate to fail the check.
⚠️ Composition problems that simplify exp and log are a favourite. The answer must contain neither exponentials nor logarithms.
True or false: ln(ab) = ln a · ln b. ___
Fill in the blank: e^(ln 5) = ___.
True or false: if ln x = −3, then x = e³. ___
Fill in the blank: an extraneous solution arises when you ___ both sides of an equation.
True or false: −ln(a/b) = ln(b/a). ___
Answers: 1. False (ln(ab) = ln a + ln b). 2. 5. 3. False (x = e⁻³). 4. Square (or perform a non-reversible operation). 5. True.
Q: Write ln(x³) − ln(x) + ln(y²) as a single logarithm.
A: ln(x³) − ln(x) + ln(y²) = ln(x³/x) + ln(y²) = ln(x²) + ln(y²) = ln(x²y²).
Q: Solve e^(2x) = 7.
A: Take ln of both sides: 2x = ln 7, so x = (ln 7)/2.
Q: Solve ln(x − 1) + ln(x + 3) = ln 5. Check for extraneous solutions.
A: Combine: ln((x − 1)(x + 3)) = ln 5. Therefore (x − 1)(x + 3) = 5, giving x² + 2x − 3 = 5, so x² + 2x − 8 = 0, which factors as (x + 4)(x − 2) = 0. Candidates: x = −4 and x = 2. Check: x = −4 makes ln(x − 1) = ln(−5), which is undefined. Extraneous. The only solution is x = 2.
Q: Simplify e^(3 ln 2).
A: 3 ln 2 = ln(2³) = ln 8. So e^(3 ln 2) = e^(ln 8) = 8.
Logarithmic and exponential equations feed directly into the calculus of growth and decay, where you differentiate and integrate functions like eᵏˣ and ln x.
The composition ideas here reappear in the chain rule, where recognising the "inner" and "outer" function is the critical step.
Logistic equations model real-world bounded growth (population dynamics, spread of disease, market saturation). This is the mathematical reason they appear on a precalculus exam.
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