Laws of logarithms — product, quotient, power rules — proofs: Worked examples

Before anything, the three tools we lean on (proved in the parent). If any symbol here feels unfamiliar, that means: a log is just a question about exponents — $lo g_{b} x$ asks "what power of $b$ gives $x$ ?".

The scenario matrix

Every log-laws question is really one of these cells. We work at least one example per cell below.

#	Cell (scenario class)	What makes it distinct	Example
A	Combine many logs into one	mix of $+, -$ and coefficients	Ex 1
B	Split one log into many	reverse direction, a product+power inside	Ex 2
C	Evaluate exactly (no calculator)	argument collapses to a clean power of the base	Ex 3
D	Negative coefficient / reciprocal	power rule with $k = - 1$ , a " $1/ M$ " appears	Ex 4
E	Fractional power ( $k$ not an integer)	roots become $lo g$ times a fraction	Ex 5
F	Solve an equation with a rejected root	domain filter throws one answer away	Ex 6
G	Degenerate / trap inputs	$lo g_{b} 1 = 0$ , $lo g_{b} b = 1$ , argument $= 0$ or $< 0$	Ex 7
H	Change-of-base twist (ratio of logs)	$\frac{l o g M}{l o g N}$ is NOT a subtraction	Ex 8
I	Word problem (real-world growth)	translate story → equation → solve	Ex 9
J	Exam-style algebraic twist	unknown base, quadratic-in-log	Ex 10

We use base-10 numeric values only where a calculator check is needed; everything else stays exact.

Cell A — Combine into one log

Worked example Ex 1 — combine

Write $\frac{1}{2} lo g_{2} x + 3 lo g_{2} y - lo g_{2} z$ as a single logarithm.

Forecast: guess — will the $\frac{1}{2}$ become a square root or a square? Will $z$ end up on top or bottom?

Step 1. $\frac{1}{2} lo g_{2} x = lo g_{2} x^{1/2} = lo g_{2} x$ . Why this step? Power rule in reverse — a number in front of a log climbs up to become an exponent. Half means square root.

Step 2. $3 lo g_{2} y = lo g_{2} y^{3}$ . Same reverse power rule.

Step 3. Add the first two: $lo g_{2} x + lo g_{2} y^{3} = lo g_{2} (y^{3} x)$ . Why this step? Sum of logs → product (product rule).

Step 4. Subtract the last: $lo g_{2} (y^{3} x) - lo g_{2} z = lo g_{2} (\frac{y ^{3} x}{z})$ . Why this step? Difference → quotient. The subtracted log lands in the denominator.

Verify: put $x = 4, y = 1, z = 2$ . Original: $\frac{1}{2} lo g_{2} 4 + 0 - lo g_{2} 2 = \frac{1}{2} (2) - 1 = 0$ . Combined: $lo g_{2} (\frac{1 \cdot 2}{2}) = lo g_{2} 1 = 0$ . ✅ Match.

Cell B — Split one log apart

Worked example Ex 2 — split (expand)

Expand $lo g_{5} (\frac{25 a ^{4}}{b})$ into separate logs.

Forecast: the $25$ is a power of $5$ — will it survive as a number or collapse to a plain integer?

Step 1. Quotient first: $lo g_{5} (25 a^{4}) - lo g_{5} b$ . Why this step? A single fraction inside → subtraction (quotient rule), separating top from bottom.

Step 2. Product on top: $lo g_{5} 25 + lo g_{5} a^{4} - lo g_{5} b$ . Why this step? $25 a^{4}$ is a product → sum.

Step 3. Power rule on each: $lo g_{5} 25 + 4 lo g_{5} a - \frac{1}{2} lo g_{5} b$ . Why this step? Exponents drop out front; $b = b^{1/2}$ .

Step 4. Simplify the pure number: $lo g_{5} 25 = lo g_{5} 5^{2} = 2$ . Why this step? $25$ is $5^{2}$ , and $lo g_{b} (b^{k}) = k$ evaluates exactly.

Result: $2 + 4 lo g_{5} a - \frac{1}{2} lo g_{5} b$ .

Verify: set $a = 1, b = 1$ : expression $= 2 + 0 - 0 = 2$ . Direct: $lo g_{5} (25 \cdot 1/1) = lo g_{5} 25 = 2$ . ✅

Cell C — Evaluate exactly

Cell D — Negative coefficient / reciprocal

Worked example Ex 4 — reciprocal via

$k = - 1$ Simplify $lo g_{7} 2 - lo g_{7} \frac{1}{2}$ .

Forecast: a reciprocal appears. Will the two terms cancel to $0$ , or reinforce each other?

Step 1. Rewrite the second term: $lo g_{7} \frac{1}{2} = lo g_{7} 2^{- 1} = - lo g_{7} 2$ . Why this step? Power rule with $k = - 1$ — a reciprocal is just "to the power $- 1$ ", so its log is the negative of the original log. Key: they do not cancel; subtracting a negative reinforces.

Step 2. Substitute: $lo g_{7} 2 - (- lo g_{7} 2) = 2 lo g_{7} 2$ . Why this step? Minus a minus is plus.

Step 3. $2 lo g_{7} 2 = lo g_{7} 2^{2} = lo g_{7} 4$ . Why this step? Power rule forward, to present one tidy log.

Answer: $2 lo g_{7} 2$ (equivalently $lo g_{7} 4$ ).

Verify: numerically $lo g_{7} 2 = ln 2/ ln 7 \approx 0.35621$ , so answer $\approx 0.71241$ . Direct: $lo g_{7} 2 - lo g_{7} 0.5 \approx 0.35621 - (- 0.35621) = 0.71241$ . ✅

Cell E — Fractional power (roots)

Cell F — Solve with a rejected root

Worked example Ex 6 — equation, domain filter

Solve $lo g_{2} (x + 6) + lo g_{2} (x) = 4$ .

Forecast: a product will build a quadratic. Guess: will both roots be valid, or does one break a domain rule?

Step 1. Product rule: $lo g_{2} (x (x + 6)) = 4$ . Why this step? Sum of logs → single log, so we can undo it in one move.

Step 2. Undo the log: $x (x + 6) = 2^{4} = 16$ . Why this step? $lo g_{2} A = 4 \Leftrightarrow A = 2^{4}$ (definition of log).

Step 3. Expand and solve: $x^{2} + 6 x - 16 = 0 \Rightarrow (x + 8) (x - 2) = 0 \Rightarrow x = - 8$ or $x = 2$ . Why this step? Standard quadratic — factor pairs of $- 16$ summing to $6$ are $8, - 2$ .

Step 4. Domain check. We need $x > 0$ and $x + 6 > 0$ . $x = - 8$ makes $lo g_{2} (x)$ undefined → reject. $x = 2$ : both $2 > 0$ and $8 > 0$ → keep. Why this step? Every logged argument must be strictly positive; algebra alone can smuggle in illegal roots.

Answer: $x = 2$ .

Verify: $lo g_{2} (8) + lo g_{2} (2) = 3 + 1 = 4$ . ✅

Cell G — Degenerate & trap inputs

Figure — Laws of logarithms — product, quotient, power rules — proofs

Worked example Ex 7 — the special values and a trap

Evaluate each, and say which is undefined: (a) $lo g_{9} 1$ (b) $lo g_{9} 9$ (c) $lo g_{9} 3$ (d) $lo g_{9} 0$ (e) $lo g_{9} (- 3)$ .

Forecast: two of these are "free" values every log has; one is a genuine fraction; two are illegal. Guess which two are illegal before reading on.

Step 1. $lo g_{9} 1 = 0$ . Why this step? $9^{0} = 1$ — any base to the power $0$ is $1$ , so the log of $1$ is always $0$ . (Red dot at the crossing in the figure.)

Step 2. $lo g_{9} 9 = 1$ . Why this step? $9^{1} = 9$ ; the log of the base itself is always $1$ .

Step 3. $lo g_{9} 3 = \frac{1}{2}$ . Why this step? $9 = 3^{2}$ , so $9^{1/2} = 3$ . Not an integer — perfectly legal, just fractional. This is the honest fraction case.

Step 4. $lo g_{9} 0$ is undefined. Why this step? No power of $9$ gives $0$ : as the exponent $\to - \infty$ , $9^{y} \to 0$ but never reaches it (see the curve diving toward $x = 0$ , the yellow asymptote). The graph never touches the vertical axis.

Step 5. $lo g_{9} (- 3)$ is undefined. Why this step? A positive base to any real power is always positive — the whole curve sits to the right of $x = 0$ , so negative inputs have no real log.

Verify: $9^{0} = 1, 9^{1} = 9, 9^{1/2} = 3$ . ✅ And $9^{y} > 0$ for every real $y$ , confirming (d),(e) have no solution.

Cell H — Change-of-base twist (the ratio trap)

Worked example Ex 8 — ratio of logs, NOT a subtraction

A student writes $\frac{lo g 8}{lo g 2} = lo g 8 - lo g 2$ . Find the true value of $\frac{lo g 8}{lo g 2}$ and expose the error.

Forecast: guess the true value — $8$ is $2^{3}$ , so the answer is a small whole number.

Step 1. Recognise the pattern: $\frac{lo g M}{lo g N} = lo g_{N} M$ (change-of-base, base-independent). Why this step? A ratio of logs is a change of base, a completely different operation from $lo g (M / N) = lo g M - lo g N$ . Division inside one log gives subtraction; division of two separate logs does not.

Step 2. So $\frac{lo g 8}{lo g 2} = lo g_{2} 8 = 3$ (since $2^{3} = 8$ ).

Step 3. The student's claim gives $lo g 8 - lo g 2 = lo g 4 = lo g_{10} 4 \approx 0.602$ — nowhere near $3$ . Why this step? Shows the two expressions are not equal, so the rule is wrong.

Answer: $3$ ; the subtraction identity does not apply to a ratio of logs.

Verify: $lo g_{10} 8/ lo g_{10} 2 = 0.90309/0.30103 = 3.0000$ . And $lo g_{10} 8 - lo g_{10} 2 = 0.60206 \neq = 3$ . ✅

Cell I — Word problem (real-world growth)

Cell J — Exam-style algebraic twist (quadratic in a log)

Worked example Ex 10 — unknown base + hidden quadratic

Solve $(lo g_{3} x)^{2} - lo g_{3} (x^{3}) + 2 = 0$ .

Forecast: the squared log looks scary, but a substitution turns it into a plain quadratic. Guess: two answers, both valid?

Step 1. Fix the middle term with the power rule: $lo g_{3} (x^{3}) = 3 lo g_{3} x$ . Why this step? Bring the exponent out front so every term is written in the single quantity $lo g_{3} x$ . Warning: $(lo g_{3} x)^{2}$ is the whole log squared — the power rule does not touch it.

Step 2. Let $u = lo g_{3} x$ . Equation becomes $u^{2} - 3 u + 2 = 0$ . Why this step? Naming the repeated expression reveals a hidden quadratic.

Step 3. Factor: $(u - 1) (u - 2) = 0 \Rightarrow u = 1$ or $u = 2$ .

Step 4. Undo the substitution. $u = 1$ : $lo g_{3} x = 1 \Rightarrow x = 3^{1} = 3$ . $u = 2$ : $lo g_{3} x = 2 \Rightarrow x = 3^{2} = 9$ . Why this step? Convert each log value back to $x$ via the definition.

Step 5. Domain check: both $x = 3$ and $x = 9$ are $> 0$ → both valid.

Answer: $x = 3$ or $x = 9$ .

Verify (x=9): $(lo g_{3} 9)^{2} - lo g_{3} (9^{3}) + 2 = 2^{2} - lo g_{3} 729 + 2 = 4 - 6 + 2 = 0$ (since $729 = 3^{6}$ ). ✅ (x=3): $1^{2} - lo g_{3} 27 + 2 = 1 - 3 + 2 = 0$ . ✅

Recall

Recall When does an algebraic solution to a log equation get rejected?

When it makes any logged argument $\leq 0$ — every argument must be strictly positive.

Recall Is

$\frac{lo g M}{lo g N}$ a subtraction? No — it is change of base, $lo g_{N} M$ . Only a quotient inside one log becomes subtraction.

Recall Why is

$lo g_{b} 0$ undefined? No real power of a positive base equals $0$ ; the curve approaches $x = 0$ but never reaches it.

Recall How do you handle

$(lo g_{b} x)^{2}$ in an equation? Substitute $u = lo g_{b} x$ to expose a quadratic; the power rule does not simplify the whole-log square.

Connections

Parent: proofs of the three laws
Exponentials — laws of indices — the index rules powering every step here.
Change of base formula — the tool behind Cell H's ratio trap.
Solving exponential equations with logs — Cells F, I, J live here.
Natural logarithm ln and e — same laws with base $e$ .
Graphs of logarithmic functions — the picture behind Cell G's domain.