Checking the Rule with Numbers

Show that $\log_2\left(\dfrac{32}{4}\right) = \log_2 32 - \log_2 4$.

Evaluate the left side directly: $\dfrac{32}{4} = 8$ and $2^3 = 8$ = $\log_2 8 = 3$

Evaluate each logarithm on the right side: $2^5 = 32$ and $2^2 = 4$ = $\log_2 32 = 5$, $\log_2 4 = 2$

Subtract: $5 - 2 = 3$ = $3$

Compare both sides: Left side $3$, right side $3$ = Both sides are equal

Answer: Both sides equal $3$, so $\log_2\left(\dfrac{32}{4}\right) = \log_2 32 - \log_2 4$.

Condensing a Difference

Write $\log_3 54 - \log_3 2$ as a single logarithm and evaluate it.

Check that the bases match: Both logarithms have base $3$ = The quotient rule applies

Combine into one logarithm (numerator: the first term): $\log_3 54 - \log_3 2 = \log_3\left(\dfrac{54}{2}\right)$ = $\log_3 27$

Evaluate: $3^3 = 27$ = $\log_3 27 = 3$

Answer: $\log_3 54 - \log_3 2 = \log_3 27 = 3$

Expanding a Quotient with a Variable

Expand $\log_5\left(\dfrac{x}{25}\right)$ for $x > 0$ and simplify as far as possible.

Apply the quotient rule: Numerator minus denominator: $\log_5\left(\dfrac{x}{25}\right) = \log_5 x - \log_5 25$ = $\log_5 x - \log_5 25$

Evaluate the number part: $5^2 = 25$ = $\log_5 25 = 2$

Write the simplified form: $\log_5 x$ cannot be simplified further = $\log_5 x - 2$

Answer: $\log_5\left(\dfrac{x}{25}\right) = \log_5 x - 2$

Quotient Rule and Product Rule Together

Expand $\log\left(\dfrac{ab}{c}\right)$ for $a, b, c > 0$.

Split the quotient first: $\log\left(\dfrac{ab}{c}\right) = \log(ab) - \log c$ = $\log(ab) - \log c$

Split the product with the product rule: $\log(ab) = \log a + \log b$ = $\log a + \log b - \log c$

Check with numbers ($a = 20$, $b = 5$, $c = 10$): $\log\left(\dfrac{100}{10}\right) = \log 10 = 1$ and $\log 20 + \log 5 - \log 10 = \log 100 - 1 = 1$ = Both give $1$

Answer: $\log\left(\dfrac{ab}{c}\right) = \log a + \log b - \log c$

Condensing Three Logarithms

Evaluate $\log 40 + \log 50 - \log 2$ without a calculator ($\log$ is the common logarithm, base 10).

Combine the sum with the product rule: $\log 40 + \log 50 = \log(40 \times 50)$ = $\log 2000$

Combine the difference with the quotient rule: $\log 2000 - \log 2 = \log\left(\dfrac{2000}{2}\right)$ = $\log 1000$

Write $1000$ as a power of $10$: $1000 = 10^3$ = $\log 1000 = 3$

Answer: $\log 40 + \log 50 - \log 2 = \log 1000 = 3$

Mistake: Writing $\dfrac{\log M}{\log N} = \log\left(\dfrac{M}{N}\right)$

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

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

Mistake: Writing $\log(M - N) = \log M - \log N$

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

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

Mistake: Subtracting in the wrong order: $\log_b\left(\dfrac{M}{N}\right) = \log_b N - \log_b M$

Why: Subtraction is not commutative, so the order matters. $\log_2\left(\dfrac{8}{2}\right) = \log_2 4 = 2$, but $\log_2 2 - \log_2 8 = 1 - 3 = -2$.

Correct: Numerator first, minus denominator: $\log_b\left(\dfrac{M}{N}\right) = \log_b M - \log_b N$.

Mistake: Writing $\log\left(\dfrac{M}{NP}\right) = \log M - \log N + \log P$

Why: Both $N$ and $P$ are in the denominator, so both logarithms are subtracted. The $+$ treats $P$ as if it were in the numerator.

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

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.

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

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.

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

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 $M_1 - M_2 = \log\left(\dfrac{A_1}{A_2}\right)$.

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

Q: Do both logarithms need the same base?

A: Yes. $\log_b M - \log_b N = \log_b\left(\dfrac{M}{N}\right)$ only works when both terms have the same base $b$. A difference like $\log_2 8 - \log_3 9$ cannot be condensed into one logarithm with this rule.

Q: What happens when the quotient is less than 1?

A: Then the logarithm is negative (for a base greater than 1), and the rule shows why: $\log_2\left(\dfrac{1}{8}\right) = \log_2 1 - \log_2 8 = 0 - 3 = -3$. In general $\log_b\left(\dfrac{1}{N}\right) = -\log_b N$.

Q: Can I expand $\log\left(\dfrac{x}{x - 2}\right)$?

A: Yes, for $x > 2$: $\log x - \log(x - 2)$. But $\log(x - 2)$ stays as it is, because there is no rule for the logarithm of a difference.

Q: Why is it called the twin of the product rule?

A: Dividing by $N$ is multiplying by $\dfrac{1}{N}$, so $\log\left(\dfrac{M}{N}\right) = \log M + \log\left(\dfrac{1}{N}\right) = \log M - \log N$. Every quotient rule step is a product rule step in disguise.