Why a Product Rule Is Useful
Logarithms turn questions about powers into questions about exponents. A useful consequence is that multiplying positive quantities inside a logarithm corresponds to adding their logarithms. This can make a product easier to evaluate or interpret when its factors are already understood separately.
The relationship is not an extra rule to memorize without explanation. It follows directly from the exponent law for multiplying powers with the same base. We will derive it carefully, state its conditions, and then use it in examples.
Logarithms and Their Conditions
For a base \(b>0\) with \(b\ne 1\), the logarithm \(\log_b x\) is defined for \(x>0\). It is the exponent to which \(b\) must be raised to obtain \(x\). In other words, \(\log_b x=r\) means \(b^r=x\).
The restrictions matter. A positive base different from \(1\) gives an exponential function that is one-to-one, so each positive input \(x\) corresponds to exactly one exponent. The input to a real logarithm must be positive. The base may be greater than \(1\) or between \(0\) and \(1\); either way, the exponential function is one-to-one and the logarithm is defined on positive inputs.
For the product rule, both factors must be positive. If \(x>0\) and \(y>0\), then \(xy>0\), so \(\log_b(xy)\), \(\log_b x\), and \(\log_b y\) are all defined.
Deriving the Product Rule
Let \(x\) and \(y\) be positive, and choose a valid base \(b\). Write each factor as a power of \(b\), using its logarithm as the exponent. Then the exponent law for multiplying powers combines those exponents into a sum.
Set \(u=\log_b x\) and \(v=\log_b y\). By the definition of logarithm, \(x=b^u\) and \(y=b^v\).
Using \(b^u b^v=b^{u+v}\), obtain \(xy=b^{u+v}\).
Since \(xy=b^{u+v}\), the definition of logarithm gives \(\log_b(xy)=u+v\).
Substituting the definitions of \(u\) and \(v\) gives the product rule.
Here is the same derivation as a chain of equalities. It also shows explicitly why the conclusion follows from the exponent law.
The last step uses the fact that if \(xy=b^{u+v}\), then the exponent that produces \(xy\) from base \(b\) is \(u+v\). That exponent is unique because \(b>0\) and \(b\ne1\). This completes the justification; the rule is a direct consequence of how logarithms undo exponentiation.
Worked Examples
Worked Example: Evaluate a Product with Base 4
Evaluate \(\log_4(16\cdot 4)\) using the product rule, then check the result directly.
Both factors are positive, and \(4>0\) with \(4\ne1\), so the product rule applies:
For the check, multiply the factors: \(16\cdot4=64\). Since \(4^3=64\), the definition gives \(\log_4 64=3\), as obtained by the product rule.
Worked Example: Use a Fractional Exponent
Evaluate \(\log_2\left(\frac{1}{8}\cdot32\right)\) using the product rule.
Each factor is positive, and both can be written as powers of \(2\): \(\frac18=2^{-3}\) and \(32=2^5\). Therefore, \(\log_2\frac18=-3\) and \(\log_2 32=5\). Apply the rule:
Check by multiplying the factors: \(\frac18\cdot32=4\). Since \(2^2=4\), the logarithm of the product is indeed \(2\).
Worked Example: Combine Two Known Logarithms
Suppose \(A=10^{0.3}\) and \(B=10^{0.7}\). Find \(\log_{10}(AB)\).
Both \(A\) and \(B\) are positive because they are powers of \(10\). By the definition of logarithm, \(\log_{10}A=0.3\) and \(\log_{10}B=0.7\). The product rule gives:
To check, multiply the powers using the exponent law: \(AB=10^{0.3}10^{0.7}=10^{0.3+0.7}=10^1=10\). Thus \(\log_{10}(AB)=\log_{10}(10)=1\), in agreement with the result.
Using the Rule with Positive Quantities
The product rule can be used whenever a positive quantity is expressed as a product. The factors do not have to be integers, and their logarithms do not have to be whole numbers. For example, if a model gives a positive quantity as \(P=ab\), then for a valid base \(b_0\), the rule says \(\log_{b_0}P=\log_{b_0}a+\log_{b_0}b\), provided \(a>0\) and \(b>0\). The base is labelled \(b_0\) here to distinguish it from a product factor.
This is useful when the factors are known separately or when their logarithms are simpler to work with than the product itself. If the logarithms of two positive factors are \(r\) and \(s\), then the logarithm of their product is \(r+s\). The rule does not require you to multiply the factors first.
For instance, if positive quantities \(M\) and \(N\) satisfy \(\log_5 M=2.4\) and \(\log_5 N=-0.4\), then:
The negative logarithm of \(N\) causes no problem: \(N\) is still positive. A positive number less than \(1\) has a negative logarithm when the base is greater than \(1\). What matters for applying this rule is positivity of the factors, not positivity of their logarithms.
What the Rule Does and Does Not Say
The product inside the logarithm becomes a sum outside it. The right-hand side is addition, not multiplication. The factors must each have defined logarithms in the real-number setting, so each factor must be positive.
| Expression | How to handle it |
|---|---|
| \(\log_b(xy)\), with \(x,y>0\) | Write \(\log_b x+\log_b y\). |
| \(\log_b x+\log_b y\), with \(x,y>0\) | It equals \(\log_b(xy)\). |
| \(\log_b(xy)\), where a factor is nonpositive | The product rule cannot split it into real logarithms of those factors. |
The condition on the individual factors is important even if their product happens to be positive. For example, \((-2)(-3)=6\), so \(\log_2((-2)(-3))\) is defined. But \(\log_2(-2)\) and \(\log_2(-3)\) are not real logarithms, so this expression cannot be split into their sum.
A Common Mistake
A frequent error is to multiply the logarithms instead of adding them. Here is a concrete check that shows the error:
The values \(4\) and \(3\) are different. The correction is to add the separate logarithms:
It is also a mistake to apply the rule when a factor is zero or negative. For example, \(\log_2(0\cdot5)\) is undefined because its input is \(0\), and the separate term \(\log_2 0\) is undefined as well. Before using the rule, check that the base is positive and not \(1\), and that both factors are positive.
Summary
The product rule for logarithms is derived from the exponent law \(b^u b^v=b^{u+v}\). For a base \(b>0\) with \(b\ne1\), and positive factors \(x\) and \(y\), the rule is:
When applying it, check the conditions first, then evaluate or retain each logarithm as appropriate, and add the results. Keeping the operation as addition is essential: multiplying the logarithms is not the product rule.
Check Your Understanding
Answer these questions using the base conditions and the product rule.
- State the conditions on \(b\), \(x\), and \(y\) needed for \(\log_b(xy)=\log_b x+\log_b y\) to hold for real logarithms.
- Evaluate \(\log_3(9\cdot27)\) using the product rule, and check by multiplying the factors.
- If \(C,D>0\), \(\log_7 C=-1\), and \(\log_7 D=4\), find \(\log_7(CD)\).
- Why does \(\log_2((-4)(-2))\) have a real value, while splitting it as \(\log_2(-4)+\log_2(-2)\) is not valid?
- For positive \(x\) and \(y\), what operation combines \(\log_b x\) and \(\log_b y\) in the product rule?
