Quotient and Power Rules for Logarithms

Why Quotients and Powers Have Simple Logarithm Rules

Logarithms are defined as inverses of exponential functions. That inverse relationship explains why multiplication becomes addition under a logarithm, as established by the product rule, and it also leads to two further useful rules: a quotient becomes a difference, and a power becomes a factor.

These rules are useful only when their inputs satisfy the definition of a real logarithm. In particular, the argument of a real logarithm must be positive. Keeping that condition visible is part of using the rules correctly, not an optional check after the calculation.

Logarithm Conditions and Inverse Relationships

For a real logarithm with base \(b\), the base must satisfy \(b>0\) and \(b\ne 1\). The argument must be positive. The notation \(\log_b(x)\) means the exponent to which \(b\) must be raised to obtain \(x\).

Definition: If \(b>0\), \(b\ne 1\), and \(x>0\), then \(\log_b(x)\) is the unique real number \(u\) such that \(b^u=x\). Equivalently, \(b^{\log_b(x)}=x\) and \(\log_b(b^u)=u\).

For example, \(\log_2(8)=3\) because \(2^3=8\). The inverse relationship works in either direction: a logarithm turns a positive number into an exponent, while an exponential turns that exponent back into the number.

We will use the standard exponent laws, including \(b^u/b^v=b^{u-v}\) and \((b^u)^r=b^{ur}\). For the power rule, \(r\) may be any real number, and \(x^r\) is understood as the real power of a positive base \(x\).

The Quotient Rule

Suppose \(x>0\) and \(y>0\), and let \(u=\log_b(x)\) and \(v=\log_b(y)\). By the definition of logarithm, \(x=b^u\) and \(y=b^v\). The quotient of these positive numbers can therefore be written as a quotient of powers with the same base.

\frac{x}{y} = \frac{b^u}{b^v} = b^{u-v}.

Taking \(\log_b\) of both sides recovers the exponent \(u-v\). Substituting the definitions of \(u\) and \(v\) gives the quotient rule.

Theorem: If \(b>0\), \(b\ne 1\), \(x>0\), and \(y>0\), then

\log_b\left(\frac{x}{y}\right) = \log_b(x)-\log_b(y).

The positivity conditions ensure that both logarithms on the right are defined. They also make \(x/y\) positive, so the logarithm on the left is defined.

The direction of the subtraction matters: the logarithm of the numerator comes first, followed by the logarithm of the denominator. This order matches the exponent law \(b^u/b^v=b^{u-v}\).

Deriving the Quotient Rule

The argument above is a complete derivation. Its central step is not a special property of logarithms introduced without explanation; it is the exponent law for dividing two powers with the same base. The logarithm then recovers the resulting exponent.

Represent each positive input as a power of the base.
Set \(u=\log_b(x)\) and \(v=\log_b(y)\), so \(x=b^u\) and \(y=b^v\).

Divide the powers.
The exponent law gives \(x/y=b^u/b^v=b^{u-v}\).

Apply the inverse relationship.
Taking \(\log_b\) of \(b^{u-v}\) gives \(u-v\), which is \(\log_b(x)-\log_b(y)\).

Worked Example: Evaluate a Quotient Logarithm

Evaluate \(\log_2(32/4)\), using the quotient rule and checking the result directly.

First, check the conditions: the base is \(2>0\) and \(2\ne 1\), while both quotient inputs, \(32\) and \(4\), are positive. The rule applies.

\log_2\left(\frac{32}{4}\right) = \log_2(32)-\log_2(4) = 5-2 = 3.

For a direct check, \(32/4=8\), and \(2^3=8\). Therefore \(\log_2(32/4)=\log_2(8)=3\), agreeing with the rule.

Worked Example: A Quotient Smaller Than One

Evaluate \(\log_5\big((1/25)/5\big)\). Both inputs in the quotient are positive, so the quotient rule applies.

\log_5\left(\frac{1/25}{5}\right) = \log_5\left(\frac{1}{25}\right)-\log_5(5) = -2-1 = -3.

To check, \((1/25)/5=1/125\), and \(5^{-3}=1/125\). Thus the logarithm of the quotient is \(-3\), as calculated.

The Power Rule

Let \(x>0\), and write \(u=\log_b(x)\), so \(x=b^u\). Raising both sides to a real power \(r\) gives \(x^r=(b^u)^r\). The exponent law for a power of a power gives \((b^u)^r=b^{ur}\). Applying \(\log_b\) recovers the exponent \(ur\), which equals \(r\log_b(x)\).

Theorem: If \(b>0\), \(b\ne 1\), \(x>0\), and \(r\in\mathbb{R}\), then

\log_b(x^r)=r\log_b(x).

Here \(x>0\) ensures that \(\log_b(x)\) is defined and that the real power \(x^r\) is positive, so the logarithm on the left is also defined.

The exponent \(r\) does not have to be an integer. It may be positive, negative, or fractional. Positivity of \(x\) is what allows the rule to be stated for every real \(r\) using real logarithms and real powers.

Rewrite the positive input as an exponential.
Let \(u=\log_b(x)\), so \(x=b^u\).

Raise both sides to the power \(r\).
Then \(x^r=(b^u)^r=b^{ur}\).

Use the inverse relationship.
It follows that \(\log_b(x^r)=ur=r\log_b(x)\).

Worked Example: A Fractional Power

Evaluate \(\log_4(8^{2/3})\) using the power rule, then verify the result by evaluating the power first.

The base \(4\) is positive and not \(1\), and \(8>0\), so the logarithm and power rule are valid. Since \(4^{3/2}=8\), we have \(\log_4(8)=3/2\). Therefore:

\log_4\left(8^{2/3}\right) = \frac{2}{3}\log_4(8) = \frac{2}{3}\cdot\frac{3}{2} = 1.

For the direct check, \(8^{2/3}=(\sqrt[3]{8})^2=2^2=4\), and \(\log_4(4)=1\). The direct calculation agrees.

Worked Example: A Negative Exponent

Evaluate \(\log_3\big((1/27)^{-2/3}\big)\). The input \(1/27\) is positive, so a negative real exponent is permitted and the power rule applies.

Since \(3^{-3}=1/27\), we have \(\log_3(1/27)=-3\). Applying the rule gives:

\log_3\left(\left(\frac{1}{27}\right)^{-2/3}\right) = -\frac{2}{3}\log_3\left(\frac{1}{27}\right) = -\frac{2}{3}\cdot(-3) = 2.

For a direct check, \((1/27)^{-2/3}=(3^{-3})^{-2/3}=3^2=9\), and \(\log_3(9)=2\). The result is verified.

Conditions and Common Mistakes

Both rules depend on the positivity of the inputs that appear inside the logarithms. The quotient rule requires \(x>0\) and \(y>0\), not merely that the quotient \(x/y\) be positive. The power rule requires \(x>0\); the rule as stated does not allow a negative base just because a particular integer power happens to be positive.

Worked Example: Why Positivity Must Be Checked

Consider \(\log_2((-4)/(-2))\). The quotient itself is positive, so the original logarithm is defined. Evaluating it directly gives \((-4)/(-2)=2\), and therefore \(\log_2(2)=1\).

A tempting but invalid step is to apply the quotient rule to the two negative numbers:

\log_2\left(\frac{-4}{-2}\right) \;\not=\; \log_2(-4)-\log_2(-2) \quad\text{as a real-logarithm calculation}.

The expressions \(\log_2(-4)\) and \(\log_2(-2)\) are not defined as real logarithms. The quotient rule requires each of its two inputs to be positive. The correction is to evaluate the quotient first: \(\log_2(2)=1\). This value is valid, but it does not come from applying the quotient rule to negative inputs.

A related mistake is to apply the power rule to a negative base without checking the conditions. For instance, \(\log_2((-2)^2)=\log_2(4)=2\) is defined, but the expression \(2\log_2(-2)\) is not defined as a real logarithm. The power rule stated here does not apply to \(x=-2\), even though the squared quantity is positive.

Key takeaway: Before applying either rule, check the base of the logarithm and the signs of the inputs inside the logarithms. A valid-looking final argument does not guarantee that every intermediate logarithm is defined.

Why These Rules Matter

The quotient and power rules make logarithms useful for understanding how multiplicative relationships behave. A quotient becomes a difference of logarithms, while a power becomes a constant factor. These identities also help evaluate logarithms exactly when the quotient or power can be recognized through exponent laws.

However, the rules are not permission to ignore domains. A reliable calculation has two parts: the algebraic step and the conditions that make the step meaningful. In particular, a real logarithm never accepts zero or a negative argument, and the separate logarithms on the right side of either rule must themselves be defined.

Summary

For a valid logarithm base \(b>0\), \(b\ne 1\), the quotient rule holds when \(x>0\) and \(y>0\), and the power rule holds when \(x>0\) and \(r\) is real.

\log_b\left(\frac{x}{y}\right) = \log_b(x)-\log_b(y), \qquad \log_b(x^r) = r\log_b(x).

Both identities follow from the definition of logarithm and the exponent laws. Applying them carefully means checking the conditions first and, when possible, verifying the result by evaluating the original quotient or power.

Check Your Understanding

Use exact values where possible, and check the domain conditions before applying a rule.

Evaluate \(\log_3(81/9)\) using the quotient rule, then verify by evaluating the quotient.
Evaluate \(\log_2(16^{-3/4})\) using the power rule, and check by evaluating the power first.
State the conditions on \(b\), \(x\), and \(y\) required for the quotient rule.
Can the real quotient rule be applied directly to \(\log_5((-25)/(-5))\)? Explain why or why not.
Why does the stated power rule require \(x>0\) when \(r\) may be any real number?