Quotient Rule for Logarithms

Learn why the logarithm of a quotient equals a difference of logarithms, and use the rule to expand and condense expressions.

Advanced25 min18 problems
Practice Problems

Definition

The quotient rule for logarithms says: the logarithm of a quotient is the logarithm of the numerator minus the logarithm of the denominator.
log⁡b(MN)=log⁡bM−log⁡bN\log_b\left(\dfrac{M}{N}\right) = \log_b M - \log_b N
The rule holds for M>0M > 0, N>0N > 0 and a base b>0b > 0 with b≠1b \neq 1.
Why it works: a logarithm is an exponent. Write log⁡bM=x\log_b M = x and log⁡bN=y\log_b N = y, so M=bxM = b^x and N=byN = b^y. Dividing powers with the same base subtracts the exponents:
MN=bxby=bx−y⇒log⁡b(MN)=x−y=log⁡bM−log⁡bN\dfrac{M}{N} = \dfrac{b^x}{b^y} = b^{x-y} \quad\Rightarrow\quad \log_b\left(\dfrac{M}{N}\right) = x - y = \log_b M - \log_b N
It is the twin of the product rule (log⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N): multiplying inside the logarithm becomes adding outside, and dividing inside becomes subtracting outside.
Example: log⁡2(324)=log⁡232−log⁡24=5−2=3\log_2\left(\dfrac{32}{4}\right) = \log_2 32 - \log_2 4 = 5 - 2 = 3, and indeed log⁡28=3\log_2 8 = 3.
You can use the rule in two directions:
  • Expand: split one logarithm of a quotient into a difference, log⁡3(x7)=log⁡3x−log⁡37\log_3\left(\dfrac{x}{7}\right) = \log_3 x - \log_3 7
  • Condense: combine a difference into one logarithm, log⁡500−log⁡5=log⁡100=2\log 500 - \log 5 = \log 100 = 2

Try it now

Which expression is equal to log⁡b(MN)\log_b\left(\dfrac{M}{N}\right) (for M,N>0M, N > 0)?

Worked Examples

Show that log⁡2(324)=log⁡232−log⁡24\log_2\left(\dfrac{32}{4}\right) = \log_2 32 - \log_2 4.
1

Evaluate the left side directly

324=8\dfrac{32}{4} = 8 and 23=82^3 = 8 → log⁡28=3\log_2 8 = 3
2

Evaluate each logarithm on the right side

25=322^5 = 32 and 22=42^2 = 4 → log⁡232=5\log_2 32 = 5, log⁡24=2\log_2 4 = 2
3

Subtract

5−2=35 - 2 = 3
4

Compare both sides

Left side 33, right side 33 → Both sides are equal

Common Mistakes

Writing log⁡Mlog⁡N=log⁡(MN)\dfrac{\log M}{\log N} = \log\left(\dfrac{M}{N}\right)

Why it's wrong: A quotient of two logarithms is not the logarithm of a quotient. Test it: log⁡100log⁡10=21=2\dfrac{\log 100}{\log 10} = \dfrac{2}{1} = 2, but log⁡(10010)=log⁡10=1\log\left(\dfrac{100}{10}\right) = \log 10 = 1.

Correct: The rule turns division INSIDE the logarithm into subtraction: log⁡(MN)=log⁡M−log⁡N\log\left(\dfrac{M}{N}\right) = \log M - \log N. The fraction log⁡Mlog⁡N\dfrac{\log M}{\log N} has no shortcut of this kind.

Writing log⁡(M−N)=log⁡M−log⁡N\log(M - N) = \log M - \log N

Why it's wrong: The rule is about a quotient inside the logarithm, not a difference. Test it: log⁡(100−10)=log⁡90≈1.954\log(100 - 10) = \log 90 \approx 1.954, but log⁡100−log⁡10=1\log 100 - \log 10 = 1.

Correct: There is no rule that splits log⁡(M−N)\log(M - N). Only log⁡(MN)\log\left(\dfrac{M}{N}\right) becomes log⁡M−log⁡N\log M - \log N.

Subtracting in the wrong order: log⁡b(MN)=log⁡bN−log⁡bM\log_b\left(\dfrac{M}{N}\right) = \log_b N - \log_b M

Why it's wrong: Subtraction is not commutative, so the order matters. log⁡2(82)=log⁡24=2\log_2\left(\dfrac{8}{2}\right) = \log_2 4 = 2, but log⁡22−log⁡28=1−3=−2\log_2 2 - \log_2 8 = 1 - 3 = -2.

Correct: Numerator first, minus denominator: log⁡b(MN)=log⁡bM−log⁡bN\log_b\left(\dfrac{M}{N}\right) = \log_b M - \log_b N.

Writing log⁡(MNP)=log⁡M−log⁡N+log⁡P\log\left(\dfrac{M}{NP}\right) = \log M - \log N + \log P

Why it's wrong: Both NN and PP are in the denominator, so both logarithms are subtracted. The ++ treats PP as if it were in the numerator.

Correct: log⁡(MNP)=log⁡M−log⁡(NP)=log⁡M−log⁡N−log⁡P\log\left(\dfrac{M}{NP}\right) = \log M - \log(NP) = \log M - \log N - \log P. Use parentheses: log⁡M−(log⁡N+log⁡P)\log M - (\log N + \log P).

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05101520-2-10123xyx = 0

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x12451020
log(x/a)-0.3010300.301030.397940.698971
log(x) - log(a)-0.3010300.301030.397940.698971
log(x) / log(a) (WRONG: not equal to log(x/a))0122.321933.321934.32193

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Practice Problems

18 problems
Problem 1 of 18
Easy
Which expression is equal to log⁡b(MN)\log_b\left(\dfrac{M}{N}\right) (for M,N>0M, N > 0)?

Why It Matters

The quotient rule turns division into subtraction, and that is exactly what many formulas need:
  • Comparing two measurements: Decibels, pH and the Richter scale all compare a quantity to a reference value with log⁡(II0)\log\left(\dfrac{I}{I_0}\right). The quotient rule shows that the difference between two levels depends only on the ratio of the two quantities.
  • Sound: Two sounds whose intensities have the ratio 10001000 differ by 10log⁡1000=3010 \log 1000 = 30 decibels, no matter how loud they are.
  • Chemistry: Two solutions whose hydrogen ion concentrations differ by a factor of 100100 differ by exactly 22 on the pH scale.
  • Simplifying expressions: Expanding log⁡(abc)\log\left(\dfrac{ab}{c}\right) into log⁡a+log⁡b−log⁡c\log a + \log b - \log c makes later algebra (and calculus) much easier.
  • Solving equations: Condensing log⁡3x−log⁡34\log_3 x - \log_3 4 into log⁡3(x4)\log_3\left(\dfrac{x}{4}\right) is often the first step in solving a logarithmic equation.
Together with the product rule and the power rule, the quotient rule lets you rewrite almost any logarithmic expression into the form you need.

Real World Applications

Comparing Sound Levels in Decibels

Sound level is $L = 10 \log\left(\dfrac{I}{I_0}\right)$ decibels, where $I$ is the intensity and $I_0$ a fixed reference. By the quotient rule, the difference of two levels is $10(\log I_A - \log I_B) = 10 \log\left(\dfrac{I_A}{I_B}\right)$: the reference $I_0$ cancels.

Example:

A sound 100 times more intense than another is 10log⁡100=2010 \log 100 = 20 decibels louder.

1Try It Yourself

A loudspeaker produces an intensity of IA=0.01I_A = 0.01 W/m², and a quiet street produces IB=0.00001I_B = 0.00001 W/m².

How many decibels louder is the loudspeaker than the street?

Step 1: Write the mathematical expression

Calculate 10log⁡(IAIB)10 \log\left(\dfrac{I_A}{I_B}\right):

Comparing Acidity with pH

Chemists define $\text{pH} = -\log[H^+]$, where $[H^+]$ is the hydrogen ion concentration. By the quotient rule, the pH difference of two solutions is the logarithm of the ratio of their concentrations.

Example:

Coffee has [H+]=10−5[H^+] = 10^{-5} and pure water 10−710^{-7}. The pH difference is log⁡(10−510−7)=log⁡100=2\log\left(\dfrac{10^{-5}}{10^{-7}}\right) = \log 100 = 2 (pH 5 versus pH 7).

2Try It Yourself

Tomato juice has [H+]=10−4[H^+] = 10^{-4} and pure water has [H+]=10−7[H^+] = 10^{-7}.

By how much is the pH of tomato juice lower than the pH of water?

Step 1: Write the mathematical expression

Calculate log⁡(10−410−7)\log\left(\dfrac{10^{-4}}{10^{-7}}\right):

Comparing Earthquakes on the Richter Scale

The Richter magnitude is $M = \log\left(\dfrac{A}{A_0}\right)$, where $A$ is the amplitude of the ground motion and $A_0$ a reference amplitude. For two earthquakes measured at the same distance, the quotient rule gives $M_1 - M_2 = \log\left(\dfrac{A_1}{A_2}\right)$.

Example:

If one earthquake has 10 times the amplitude of another, its magnitude is log⁡10=1\log 10 = 1 higher.

3Try It Yourself

Two earthquakes are recorded at the same distance. The first has amplitude A1=50000A_1 = 50000 micrometers, the second A2=500A_2 = 500 micrometers. The second earthquake has magnitude 44.

What is the magnitude of the first earthquake?

Step 1: Write the mathematical expression

Calculate 4+log⁡(A1A2)4 + \log\left(\dfrac{A_1}{A_2}\right):

Key Takeaways

  • 1Quotient rule: log⁡b(MN)=log⁡bM−log⁡bN\log_b\left(\dfrac{M}{N}\right) = \log_b M - \log_b N for M,N>0M, N > 0
  • 2It comes from the exponent rule bxby=bx−y\dfrac{b^x}{b^y} = b^{x-y}: dividing powers subtracts exponents
  • 3Expanding splits a logarithm of a quotient into a difference; condensing combines a difference into one logarithm
  • 4Numerator first: the logarithm of the denominator is always the one subtracted
  • 5With the product rule: log⁡(abc)=log⁡a+log⁡b−log⁡c\log\left(\dfrac{ab}{c}\right) = \log a + \log b - \log c and log⁡(abc)=log⁡a−log⁡b−log⁡c\log\left(\dfrac{a}{bc}\right) = \log a - \log b - \log c
  • 6log⁡Mlog⁡N\dfrac{\log M}{\log N} and log⁡(M−N)\log(M - N) are NOT equal to log⁡(MN)\log\left(\dfrac{M}{N}\right)

Frequently Asked Questions

Yes. log⁡bM−log⁡bN=log⁡b(MN)\log_b M - \log_b N = \log_b\left(\dfrac{M}{N}\right) only works when both terms have the same base bb. A difference like log⁡28−log⁡39\log_2 8 - \log_3 9 cannot be condensed into one logarithm with this rule.
Yes. log⁡bM−log⁡bN=log⁡b(MN)\log_b M - \log_b N = \log_b\left(\dfrac{M}{N}\right) only works when both terms have the same base bb. A difference like log⁡28−log⁡39\log_2 8 - \log_3 9 cannot be condensed into one logarithm with this rule.
Then the logarithm is negative (for a base greater than 1), and the rule shows why: log⁡2(18)=log⁡21−log⁡28=0−3=−3\log_2\left(\dfrac{1}{8}\right) = \log_2 1 - \log_2 8 = 0 - 3 = -3. In general log⁡b(1N)=−log⁡bN\log_b\left(\dfrac{1}{N}\right) = -\log_b N.
Yes, for x>2x > 2: log⁡x−log⁡(x−2)\log x - \log(x - 2). But log⁡(x−2)\log(x - 2) stays as it is, because there is no rule for the logarithm of a difference.
Dividing by NN is multiplying by 1N\dfrac{1}{N}, so log⁡(MN)=log⁡M+log⁡(1N)=log⁡M−log⁡N\log\left(\dfrac{M}{N}\right) = \log M + \log\left(\dfrac{1}{N}\right) = \log M - \log N. Every quotient rule step is a product rule step in disguise.

Glossary

Quotient rule
The property log⁡b(MN)=log⁡bM−log⁡bN\log_b\left(\dfrac{M}{N}\right) = \log_b M - \log_b N for M,N>0M, N > 0
Quotient
The result of a division; in MN\dfrac{M}{N}, MM is the numerator and NN the denominator
Expand
Rewrite one logarithm of a product or quotient as a sum or difference of simpler logarithms
Condense
Combine a sum or difference of logarithms with the same base into a single logarithm
Argument
The number inside a logarithm, such as MM in log⁡bM\log_b M; it must be positive

Formula Card

Quotient rule

log⁡b(MN)=log⁡bM−log⁡bN\log_b\left(\dfrac{M}{N}\right) = \log_b M - \log_b N

For $M > 0$, $N > 0$, $b > 0$, $b \neq 1$

Reciprocal

log⁡b(1N)=−log⁡bN\log_b\left(\dfrac{1}{N}\right) = -\log_b N

Special case with $M = 1$, because $\log_b 1 = 0$

With the product rule

log⁡b(MNP)=log⁡bM+log⁡bN−log⁡bP\log_b\left(\dfrac{MN}{P}\right) = \log_b M + \log_b N - \log_b P

Numerator factors are added, denominator factors subtracted

Not a rule

log⁡bMlog⁡bN≠log⁡b(MN)\dfrac{\log_b M}{\log_b N} \neq \log_b\left(\dfrac{M}{N}\right)

A quotient of logarithms is not the logarithm of a quotient

Today's challengeOne problem, three tries

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