Why Condense Logarithms?
A logarithmic expression can contain several terms, each describing part of a multiplicative relationship. Combining those terms into one logarithm can make the structure easier to see and can simplify later calculations. The main task is to use the logarithm rules in the reverse direction from expanding: sums become products, differences become quotients, and constant multiples become exponents.
These rules have conditions. Every logarithm must have a positive argument, and the logarithms being combined must have the same base. Keeping those requirements visible is part of a correct condensation, not an optional final check.
Rules for Condensing
Let \(b\) be a positive real number with \(b\ne 1\). The logarithm \(\log_b A\) is defined when \(A>0\). For positive \(A\) and \(B\), the product and quotient rules are
For a real constant \(c\) and \(A>0\), the power rule is
Since \(A>0\), its real power \(A^c\) is positive, so the logarithm on the right is defined. A negative coefficient produces a reciprocal power; for example, \(-2\log_b A=\log_b(A^{-2})=\log_b(1/A^2)\). A fractional coefficient produces a root when the exponent is a suitable fraction, as in \(\frac12\log_b A=\log_b(\sqrt A)\).
Why the Rules Work
The rules follow from the definition of a logarithm and the laws of exponents. If \(u=\log_b A\), then \(b^u=A\); likewise, if \(v=\log_b B\), then \(b^v=B\). The exponent laws show that \(b^{u+v}=AB\). Because \(\log_b\) is the inverse of the exponential function with base \(b\), the exponent whose power of \(b\) is \(AB\) is \(\log_b(AB)\). Thus \(u+v=\log_b(AB)\).
For the quotient rule, the exponent law for subtraction gives \(b^{u-v}=b^u/b^v=A/B\). Since \(A\) and \(B\) are positive, their quotient is positive, and taking the logarithm gives the rule. For the power rule, \(b^{cu}=(b^u)^c=A^c\). The condition \(A>0\) ensures \(A^c\) is defined as a real power for every real \(c\). Therefore, \(cu=\log_b(A^c)\), which is the stated rule.
A Reliable Condensing Process
When an expression has several terms, it helps to work in a deliberate order. First check that the logarithms have a common base and record the conditions that make each original argument positive. Then move coefficients into exponents, combine sums as products, and combine differences as quotients. Finally, state the domain of the original expression alongside the condensed result.
Write down the positivity condition for every original logarithm argument. Do not discard these conditions later.
Rewrite \(c\log_b A\) as \(\log_b(A^c)\), provided \(A>0\).
Use products for sums and quotients for differences, keeping the order of numerator and denominator correct.
Expand the single logarithm using the same rules and check that the original expression returns, with the same domain restrictions.
Worked Examples
Worked Example: Condensing a Variable Expression
Condense \(2\log_3 x+\log_3(x-1)-\log_3 9\), and state its domain.
First require each original argument to be positive. The conditions are \(x>0\), \(x-1>0\), and \(9>0\). Taken together, these give \(x>1\).
Move the coefficient \(2\) into an exponent using the power rule:
Now combine the sum as a product, then combine the subtraction as a quotient:
The condensed argument is positive for \(x>1\), because \(x^2>0\), \(x-1>0\), and \(9>0\). The original domain is also \(x>1\), so the domain has been preserved.
Check by expanding the result on that domain:
This recovers the given expression and confirms the condensation.
Worked Example: Using a Fractional and a Negative Coefficient
Condense \(\log_2(3x+1)+\frac12\log_2(x-1)-2\log_2(x+2)\), and state its domain.
The original arguments must all be positive. The conditions are \(3x+1>0\), \(x-1>0\), and \(x+2>0\). These require \(x>-\frac13\), \(x>1\), and \(x>-2\), respectively. All three hold together exactly when \(x>1\).
Move the coefficients into exponents:
Combine the three logarithms. The negative exponent places the corresponding factor in the denominator:
For \(x>1\), the numerator factors are positive, and the denominator is positive and nonzero. Thus the combined argument is positive, and the condensed expression is defined throughout the original domain \(x>1\).
To check, expand the result:
where the expansion is valid for \(x>1\). This matches the starting expression.
Worked Example: Condensing and Checking Exact Values
Condense \(\log_5(25a)-3\log_5 b+\frac12\log_5 c\), then evaluate the original and condensed expressions when \(a=1\), \(b=5\), and \(c=25\).
The original logarithm arguments must satisfy \(25a>0\), \(b>0\), and \(c>0\); the given values meet all three conditions. Move the coefficients into exponents and combine:
Substitute \(a=1\), \(b=5\), and \(c=25\) into the original expression. Since \(\log_5 25=2\), \(\log_5 5=1\), and \(\frac12\log_5 25=1\), its value is
Now substitute into the condensed argument:
Both forms give \(0\). This exact check confirms the combined expression for these values, while the logarithm rules establish the identity for all positive \(a\), \(b\), and \(c\).
Common Mistakes and Domain Checks
A frequent error is to treat a sum inside a logarithm as though it followed the product rule. The product rule concerns a sum of logarithms, not the logarithm of a sum. For example, the incorrect calculation \(\log_2 1+\log_2 1=\log_2(1+1)=1\) changes the value: the left side is \(0+0=0\), while the right side is \(1\). The correct rule gives
Another important issue is preserving the original domain. Consider \(\log_2(x-1)-\log_2(x-1)+\log_2 x\). The original expression requires both \(x-1>0\) and \(x>0\), so its domain is \(x>1\). On that domain, condensing gives \(\log_2\left(\frac{(x-1)x}{x-1}\right)=\log_2 x\). Although this simplified expression is defined for every \(x>0\), the original expression is not. The condensed formula is equivalent to the original only on the original domain \(x>1\); cancellation does not extend that domain.
Finally, do not combine logarithms with different bases as if the product or quotient rule applied directly. The rules stated here require a common base. If the bases differ, first evaluate or rewrite the terms using an appropriate method rather than joining their arguments by multiplication or division.
Why Condensation Is Useful
A single logarithm can expose the multiplicative structure hidden in a sum of logarithmic terms. It can also make an expression more compact and easier to compare with another logarithm. In later calculations, this compact form may be useful for solving equations or simplifying models. Whatever the purpose, the same checks remain essential: common base, positive original arguments, correct placement of factors, and preservation of the original domain.
Check Your Understanding
For each question, keep both the logarithm rules and the domain conditions in mind.
- Condense \(\log_4 u+\log_4 v\). What conditions on \(u\) and \(v\) are required?
- Condense \(3\log_7 t-\log_7 s\), assuming both original logarithms are defined.
- Write \(-\frac12\log_3 r\) as a single logarithm. What condition on \(r\) is needed?
- Find the domain of \(\log_2(x+2)+\log_2(x-3)\), then give its condensed form on that domain.
- Explain why \(\log_5(m+n)\) cannot generally be rewritten as \(\log_5 m+\log_5 n\).
