Quotient Rule of Logarithms

The quotient rule of logarithms is a shortcut for rewriting the logarithm of a division as a subtraction of two logarithms. Instead of evaluating \(\log_b\left(\frac{x}{y}\right)\) directly, you can split it into \(\log_b x - \log_b y\), as long as \(x\), \(y\), and the base \(b\) are all positive and \(b \ne 1\). This is one of the core logarithm rules, alongside the product rule of logarithms, and it shows up constantly when you simplify expressions or solve equations that contain logs.

For any positive base \(b\) (with \(b \ne 1\)) and positive numbers \(x\) and \(y\):

In words: the log of a quotient is the log of the numerator minus the log of the denominator, provided every log in sight uses the same base.

The quotient rule of logarithms is really just the division rule for exponents wearing a different outfit. If \(a^m = x\) and \(a^n = y\), then \(\frac{a^m}{a^n} = a^{m-n} = \frac{x}{y}\). Rewriting each exponential statement in log form, using the same conversion you practiced when you learned to go from logarithm to exponential form, gives \(m = \log_b x\), \(n = \log_b y\), and \(m - n = \log_b\left(\frac{x}{y}\right)\). Substituting shows that \(\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y\), which is exactly the rule.

You can also see the rule at work on the graph of a log function. Look at \(y = \log_2 x\) below: the value at \(x = 8\) is \(3\), the value at \(x = 4\) is \(2\), and the difference \(3 - 2 = 1\) matches the value at \(x = 2 = \frac{8}{4}\).

Example 1: Numerical logs. Simplify \(\log_3\left(\frac{27}{9}\right)\) using the quotient rule.

\(\log_3\left(\frac{27}{9}\right) = \log_3 27 - \log_3 9 = 3 - 2 = 1\). You can check this directly, since \(\frac{27}{9} = 3\) and \(\log_3 3 = 1\).

Example 2: Expanding a variable expression. Expand \(\log_5\left(\frac{7x}{y}\right)\).

Apply the quotient rule first, treating \(7x\) as the numerator: \(\log_5\left(\frac{7x}{y}\right) = \log_5(7x) - \log_5 y\). Then apply the product rule of logarithms to \(\log_5(7x)\), giving \(\log_5 7 + \log_5 x - \log_5 y\) as the fully expanded form.

Example 3: Condensing a difference. Write \(\log_4 20 - \log_4 5\) as a single logarithm.

Reading the rule in reverse, a difference of logs with the same base becomes the log of a quotient: \(\log_4 20 - \log_4 5 = \log_4\left(\frac{20}{5}\right) = \log_4 4 = 1\). This "condensing" direction is exactly what you need when you solve logarithmic equations that have several log terms on one side.

Watch out for these slip-ups when applying the quotient rule of logarithms:

• Do not subtract \(x\) and \(y\) themselves, only subtract \(\log_b x\) and \(\log_b y\). \(\log_b\left(\frac{x}{y}\right)\) is not \(\log_b(x - y)\).
• Do not apply the rule to logs with different bases; \(\log_2 x - \log_3 y\) cannot be combined this way.
• Do not forget the domain restriction; the rule only applies where \(x > 0\) and \(y > 0\).
• Do not confuse this with the product rule of logarithms, which adds logs for a multiplication, not a division.

Try this one on your own before checking the solution: expand \(\log_2\left(\frac{x^3}{y}\right)\) as far as possible.

Using the quotient rule gives \(\log_2(x^3) - \log_2 y\), and since \(\log_2(x^3) = 3\log_2 x\) by the power rule, the fully expanded answer is \(3\log_2 x - \log_2 y\).

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