Laws of logarithms — product, quotient, power rules — proofs: Exercises

Intuition How to use this page

This is the practice-and-check companion to the Laws of Logarithms parent note. Every problem has a complete solution hidden inside a collapsible callout — cover it, attempt the problem yourself, then reveal. The difficulty climbs in five deliberate steps:

L1 Recognition — can you spot which rule applies?
L2 Application — can you use one rule cleanly?
L3 Analysis — can you chain several rules and simplify?
L4 Synthesis — can you solve equations built from the rules?
L5 Mastery — can you handle proofs, awkward bases, and traps?

First, the notation. Throughout this page $b$ is the base of the logarithm, and $M$ and $N$ are just placeholder names for two positive numbers (or expressions) that we are taking the log of — think " $M$ = my number, $N$ = the next number". Whenever you see $M$ or $N$ you may mentally swap in any positive value like $4$ , $8$ , or $x^{2}$ ; they are stand-ins so we can state a rule once instead of writing it out for every possible number.

With that agreed, make sure these three facts are in your bones (they are the only tools you need): $lo g_{b} (M N) = lo g_{b} M + lo g_{b} N, lo g_{b} (\frac{M}{N}) = lo g_{b} M - lo g_{b} N, lo g_{b} (M^{k}) = k lo g_{b} M$ and the two "cancel" identities $b^{l o g_{b} x} = x$ and $lo g_{b} (b^{k}) = k$ .

A quick reminder of the words behind the symbols, because we will lean on them constantly:

Before the drills, one picture to anchor what "multiply→add" really means — read it, then keep it in your mind's eye through every exercise.

Figure s01 — the exponent-ruler picture of the product rule. (A teal arrow of length 2 sits from 0 to 2 on a horizontal " $lo g_{2}$ " axis, labelled $lo g_{2} 4 = 2$ . A plum arrow of length 3 is laid end-to-end starting where teal stops, from 2 to 5, labelled $lo g_{2} 8 = 3$ . A burnt-orange arrow underneath runs the whole way, from 0 to 5, labelled $lo g_{2} (4 \times 8) = lo g_{2} 32 = 5$ , showing the two lengths add up.)

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

The teal ruler measures $lo g_{b} M$ ; the plum ruler, laid end-to-end from where teal stops, measures $lo g_{b} N$ . Their combined length is $lo g_{b} M + lo g_{b} N$ — and it lands exactly on $lo g_{b} (M N)$ . Adding lengths of the exponent-ruler = multiplying the numbers. That single image is the product rule; the quotient rule just runs the plum ruler backwards, and the power rule stacks $k$ copies of the same ruler.

Level 1 — Recognition

Goal: identify which single law turns the expression into something simpler.

L1.1

Write $lo g_{5} 7 + lo g_{5} 4$ as a single logarithm.

Recall Solution

WHAT: we see a sum of two logs with the same base $5$ . WHY the product rule: the rule $lo g_{b} M + lo g_{b} N = lo g_{b} (M N)$ reads "a sum of logs equals the log of the product". A sum triggers a multiply. $lo g_{5} 7 + lo g_{5} 4 = lo g_{5} (7 \times 4) = lo g_{5} 28$ Answer: $lo g_{5} 28$ .

L1.2

Write $lo g_{2} 30 - lo g_{2} 5$ as a single logarithm and then evaluate it.

Recall Solution

WHAT: a difference of same-base logs. WHY the quotient rule: $lo g_{b} M - lo g_{b} N = lo g_{b} (M / N)$ — a subtraction triggers a divide. $lo g_{2} 30 - lo g_{2} 5 = lo g_{2} (\frac{30}{5}) = lo g_{2} 6$ That is as far as the rule takes us. ( $6$ is not a nice power of $2$ , so $lo g_{2} 6 \approx 2.585$ is the final simplified form.) Answer: $lo g_{2} 6$ .

L1.3

Rewrite $lo g_{3} (x^{5})$ so that the power is no longer inside the log.

Recall Solution

WHAT: the argument $x^{5}$ carries an exponent sitting inside the log — that is the flag for the power rule. WHY the power rule: $lo g_{b} (M^{k}) = k lo g_{b} M$ lets an exponent on the argument step out to the front as a multiplier. $lo g_{3} (x^{5}) = 5 lo g_{3} x$ Answer: $5 lo g_{3} x$ .

Level 2 — Application

Goal: use one rule to get a clean number, no calculator.

L2.1

Evaluate $lo g_{2} 48 - lo g_{2} 3$ without a calculator.

Recall Solution

Step 1 (quotient rule). WHY: a difference of same-base logs collapses into a single log of a division, and $48/3$ simplifies nicely: $lo g_{2} 48 - lo g_{2} 3 = lo g_{2} (\frac{48}{3}) = lo g_{2} 16$ Step 2 (cancel identity). WHY: rewriting $16$ as a power of the base lets $lo g_{b} (b^{k}) = k$ finish the job. $16 = 2^{4}$ , so $lo g_{2} (2^{4}) = 4$ . Answer: $4$ .

L2.2

Evaluate $lo g_{6} 4 + lo g_{6} 9$ without a calculator.

Recall Solution

Step 1 (product rule). WHY: a sum of same-base logs becomes the log of the product, and $4 \times 9 = 36$ is a power of the base $6$ : $lo g_{6} 4 + lo g_{6} 9 = lo g_{6} (4 \times 9) = lo g_{6} 36$ Step 2 (cancel identity). WHY: $36 = 6^{2}$ , so $lo g_{6} (6^{2}) = 2$ by $lo g_{b} (b^{k}) = k$ . Answer: $2$ .

L2.3

Given $lo g_{10} 2 \approx 0.3010$ , find $lo g_{10} 8$ without a calculator.

Recall Solution

WHAT: $8$ hides an exponent, $8 = 2^{3}$ — the flag for the power rule. WHY the power rule: pulling the $3$ out front converts an unknown log into a multiple of the known value $lo g_{10} 2$ . $lo g_{10} 8 = lo g_{10} (2^{3}) = 3 lo g_{10} 2 \approx 3 (0.3010) = 0.9030$ Answer: $\approx 0.9030$ .

Level 3 — Analysis

Goal: chain several rules, forwards and in reverse.

L3.1

Write $2 lo g_{3} x + lo g_{3} 5 - lo g_{3} y$ as a single logarithm. (Here $x > 0$ and $y > 0$ so the logs exist.)

Recall Solution

Step 1 (power rule in reverse). WHY: to combine logs they must be bare logs with no coefficient — so first fold the $2$ back inside as an exponent: $2 lo g_{3} x = lo g_{3} (x^{2})$ Step 2 (product rule). WHY: the two terms are now added bare logs, and a sum of logs becomes the log of a product: $lo g_{3} (x^{2}) + lo g_{3} 5 = lo g_{3} (5 x^{2})$ Step 3 (quotient rule). WHY: the remaining term is subtracted, and a difference of logs becomes the log of a division: $lo g_{3} (5 x^{2}) - lo g_{3} y = lo g_{3} (\frac{5 x ^{2}}{y})$ Answer: $lo g_{3} (\frac{5 x ^{2}}{y})$ .

L3.2

Express $lo g_{b} (\frac{a ^{3} c}{d ^{2}})$ as a sum/difference of simple logs, with no powers or roots inside. Assume $a > 0, c > 0, d > 0$ (so every log below is defined).

Recall Solution

Step 1 (quotient rule). WHY: the argument is a fraction, and a log of a division splits into a subtraction of two logs: $lo g_{b} (a^{3} c) - lo g_{b} (d^{2})$ Step 2 (product rule). WHY: the numerator is a product $a^{3} \times c$ , and a log of a product splits into a sum: $lo g_{b} (a^{3}) + lo g_{b} (c) - lo g_{b} (d^{2})$ Step 3 (power rule everywhere). WHY: each remaining log has an exponent on its argument (write $c = c^{1/2}$ ), and the power rule brings each exponent to the front as a multiplier: $3 lo g_{b} a + \frac{1}{2} lo g_{b} c - 2 lo g_{b} d$ Answer: $3 lo g_{b} a + \frac{1}{2} lo g_{b} c - 2 lo g_{b} d$ .

L3.3

Evaluate $lo g_{2} 40 + lo g_{2} 0.8 - lo g_{2} 5$ without a calculator.

Recall Solution

Step 1. WHY: combining first lets us see whether the numbers cancel to a power of $2$ — sums become products, the difference becomes a division: $lo g_{2} (\frac{40 \times 0.8}{5}) = lo g_{2} (\frac{32}{5})$ Hmm — $32/5 = 6.4$ is not a power of $2$ . Let's recheck the arithmetic: $40 \times 0.8 = 32$ , and $32/5 = 6.4$ . So the tidy answer is $lo g_{2} 6.4$ … but let us try a smarter grouping to be sure nothing cancels: $40 \times 0.8 = 32 = 2^{5}$ , then $\div 5$ gives $32/5$ . There is genuinely a leftover $5$ . Answer: $lo g_{2} (\frac{32}{5}) = lo g_{2} 6.4 \approx 2.678$ . (Not everything collapses to an integer — recognising when it doesn't is part of the skill.)

Level 4 — Synthesis

Goal: combine the rules to solve equations — and always check the domain.

L4.1

Solve $lo g_{5} (x) + lo g_{5} (x - 4) = 1$ .

Recall Solution

Step 1 (product rule). WHY: the two logs share base $5$ and are added, so a sum of logs becomes the log of a product — collapsing two logs into one we can undo: $lo g_{5} (x (x - 4)) = 1$ . Step 2 (undo the log). WHY: by the very definition of a log, $lo g_{5} A = 1$ asks "what power of $5$ gives $A$ ?" and the answer is stated to be $1$ , so $A = 5^{1}$ . Applying the definition converts the log equation into an ordinary algebraic one: $x (x - 4) = 5$ . Step 3 (solve the quadratic). WHY: expanding gives a standard quadratic we can factor: $x^{2} - 4 x - 5 = 0 \Rightarrow (x - 5) (x + 1) = 0 \Rightarrow x = 5 or x = - 1$ . Step 4 (domain check). WHY: every logged argument must be $> 0$ , so we must test each root against the original logs. If $x = - 1$ then $lo g_{5} (x) = lo g_{5} (- 1)$ is undefined — reject. $x = 5$ gives $lo g_{5} 5 + lo g_{5} 1 = 1 + 0 = 1$ ✓. Answer: $x = 5$ .

L4.2

Solve $lo g_{2} (x + 6) - lo g_{2} (x - 1) = 3$ .

Recall Solution

Step 1 (quotient rule). WHY: the two logs share base $2$ and are subtracted, so a difference of logs becomes the log of a division — again collapsing to a single log: $lo g_{2} (\frac{x + 6}{x - 1}) = 3$ . Step 2 (undo the log). WHY: by definition $lo g_{2} A = 3$ means " $A$ is the number you get by raising $2$ to the power $3$ ", so $A = 2^{3} = 8$ : $\frac{x + 6}{x - 1} = 8$ . Step 3 (solve). WHY: clearing the fraction turns it into a linear equation we can rearrange: $x + 6 = 8 (x - 1) \Rightarrow x + 6 = 8 x - 8 \Rightarrow 14 = 7 x \Rightarrow x = 2$ . Step 4 (domain). WHY: both original arguments must be positive, so we need $x + 6 > 0$ and $x - 1 > 0$ , i.e. $x > 1$ . Since $2 > 1$ ✓. Check: $lo g_{2} 8 - lo g_{2} 1 = 3 - 0 = 3$ ✓. Answer: $x = 2$ .

L4.3

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

Recall Solution

Step 1 (power rule). WHY: the left side has a coefficient $2$ in front of a log; folding it inside as an exponent puts both sides into the same " $lo g_{3}$ of something" shape: $2 lo g_{3} x = lo g_{3} (x^{2})$ , so $lo g_{3} (x^{2}) = lo g_{3} (2 x + 3)$ . Step 2 (logs equal ⇒ arguments equal). WHY this is legal: for a fixed base with $b > 0, b \neq = 1$ the function $lo g_{b}$ is one-to-one (base $3$ here satisfies that), meaning no two different positive inputs can give the same output — so equal outputs force equal inputs: $x^{2} = 2 x + 3$ . Step 3 (solve). WHY: rearranging gives a factorable quadratic: $x^{2} - 2 x - 3 = 0 \Rightarrow (x - 3) (x + 1) = 0 \Rightarrow x = 3 or x = - 1$ . Step 4 (domain). WHY: $lo g_{3} x$ requires $x > 0$ , so we test each root: reject $x = - 1$ ; keep $x = 3$ . Check: $2 lo g_{3} 3 = 2$ , and $lo g_{3} (2 \cdot 3 + 3) = lo g_{3} 9 = 2$ ✓. Answer: $x = 3$ .

Level 5 — Mastery

Goal: proofs, unusual bases, and the subtlest traps.

L5.1

Prove the power rule $lo g_{b} (M^{k}) = k lo g_{b} M$ from the definition, for any real $k$ .

Recall Solution

Step 1 (name the log). WHY: giving the unknown log a single letter lets us treat it as an exponent and manipulate it. Let $p = lo g_{b} M$ . Step 2 (definition). WHY: by what $lo g$ means, " $p$ is the power that turns $b$ into $M$ ", so $b^{p} = M$ — this converts the log into exponential form where the index laws live. Step 3 (raise to power $k$ ). WHY: we want $M^{k}$ , so raise both sides to the $k$ ; a power of a power multiplies the exponents: $M^{k} = (b^{p})^{k} = b^{p k}$ (see Exponentials — laws of indices). Step 4 (take $lo g_{b}$ ). WHY: applying $lo g_{b}$ undoes the base cleanly via $lo g_{b} (b^{n}) = n$ : $lo g_{b} (M^{k}) = lo g_{b} (b^{p k}) = p k$ . Step 5 (replace $p$ ). WHY: substituting back $p = lo g_{b} M$ returns the statement in terms of the original quantities: $p k = k lo g_{b} M$ . $■$

L5.2

Simplify $lo g_{9} 27$ using log laws (write $9$ and $27$ as powers of $3$ ).

Recall Solution

WHY rewrite: $9$ and $27$ share the hidden base $3$ ( $9 = 3^{2}$ , $27 = 3^{3}$ ), and matching bases lets us compare exponents directly. Step 1 (name and undo). WHY: naming the log turns the problem into an exponential equation. Let $y = lo g_{9} 27$ , so by definition $9^{y} = 27$ . Step 2 (rewrite to a common base). WHY: with everything in base $3$ the two sides become directly comparable: $(3^{2})^{y} = 3^{3} \Rightarrow 3^{2 y} = 3^{3}$ . Step 3 (equate exponents). WHY: since $3^{a} = 3^{b}$ forces $a = b$ (the exponential function is one-to-one): $2 y = 3 \Rightarrow y = \frac{3}{2}$ . Answer: $lo g_{9} 27 = \frac{3}{2}$ .

L5.3

Given $lo g_{b} 2 = 0.35$ and $lo g_{b} 3 = 0.55$ , find $lo g_{b} 24$ and $lo g_{b} 1.5$ .

Recall Solution

WHY: break $24$ and $1.5$ into factors of $2$ and $3$ so the given values plug straight in via the product/quotient/power rules. Step 1 ( $24$ ). WHY: $24 = 2^{3} \times 3$ , so the product rule splits it into a sum and the power rule pulls the exponent out front: $lo g_{b} 24 = 3 lo g_{b} 2 + lo g_{b} 3 = 3 (0.35) + 0.55 = 1.05 + 0.55 = 1.60$ Step 2 ( $1.5$ ). WHY: $1.5 = \frac{3}{2}$ , so the quotient rule turns the division into a subtraction: $lo g_{b} 1.5 = lo g_{b} 3 - lo g_{b} 2 = 0.55 - 0.35 = 0.20$ Answers: $lo g_{b} 24 = 1.60$ , $lo g_{b} 1.5 = 0.20$ .

L5.4

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

Recall Solution

WHY substitute: the expression is a quadratic in the whole log, so replacing that repeated block with one letter reveals a familiar quadratic. Let $u = lo g_{2} x$ . This is where the classic trap lives — see below. Step 1 (solve the quadratic). WHY: in terms of $u$ it factors cleanly: $u^{2} - 3 u + 2 = 0 \Rightarrow (u - 1) (u - 2) = 0 \Rightarrow u = 1 or u = 2$ . Step 2 (undo the substitution). WHY: $u$ was only a nickname for $lo g_{2} x$ , so translate each value back using the definition of log: $lo g_{2} x = 1 \Rightarrow x = 2^{1} = 2$ ; and $lo g_{2} x = 2 \Rightarrow x = 2^{2} = 4$ . Step 3 (domain). WHY: $lo g_{2} x$ needs $x > 0$ ; both $2 > 0$ and $4 > 0$ ✓. Answer: $x = 2$ or $x = 4$ .

Answer key (fast check)

#	Answer	#	Answer
L1.1	$lo g_{5} 28$	L3.3	$lo g_{2} 6.4 \approx 2.678$
L1.2	$lo g_{2} 6$	L4.1	$x = 5$
L1.3	$5 lo g_{3} x$	L4.2	$x = 2$
L2.1	$4$	L4.3	$x = 3$
L2.2	$2$	L5.2	$\frac{3}{2}$
L2.3	$\approx 0.9030$	L5.3	$1.60, 0.20$
L3.1	$lo g_{3} \frac{5 x ^{2}}{y}$	L5.4	$x = 2$ or $4$
L3.2	$3 lo g_{b} a + \frac{1}{2} lo g_{b} c - 2 lo g_{b} d$

Connections

Laws of logarithms — product, quotient, power rules — proofs (index 3.2.8) — the parent note these drills belong to.
Exponentials — laws of indices — the exponent rules powering every proof.
Change of base formula — resolves the L2 "ratio of logs" trap.
Solving exponential equations with logs — the L4/L5 equation-solving toolkit.
Natural logarithm ln and e — the same laws with base $e$ .
Graphs of logarithmic functions — why the domain $x > 0$ forces us to reject phantom roots.