Why Log Laws Help Solve Equations
A logarithmic equation can contain several logarithms of expressions involving the unknown. When those logarithms have the same base, the product and quotient rules can turn a sum or difference into a single logarithm. That often makes it possible to convert the equation into an algebraic equation and solve it.
The conversion is useful, but it must be handled with care. A logarithm is defined only when its argument is positive, and combining terms does not remove that requirement. Therefore, solving the algebraic equation gives candidates; each candidate still has to be checked in the original logarithmic equation.
The Log Laws Used in Solving
For positive numbers \(A\) and \(B\), and a base \(b>0\) with \(b\ne 1\), the product and quotient rules state:
These rules require the same base in both logarithms. The positive-argument conditions matter: both \(A\) and \(B\) must be positive before either rule can be applied. Once logarithms have been combined, the argument \(AB\) or \(A/B\) is positive as well.
A second fact turns a single logarithmic equation into an equation about its argument. Since logarithms with base \(b>0\), \(b\ne1\), are one-to-one, for positive \(U\) and \(V\) we have:
Also, \(\log_b U=c\) is equivalent to \(U=b^c\), provided \(U>0\). These are the inverse-function relationships between logarithms and exponentials.
Why Combining the Logarithms Works
Consider a sum \(\log_b A+\log_b B\), where \(A>0\), \(B>0\), and \(b>0\), \(b\ne1\). Write \(p=\log_b A\) and \(q=\log_b B\). By the definition of logarithm, \(A=b^p\) and \(B=b^q\). Therefore \(AB=b^p b^q=b^{p+q}\). Taking the logarithm base \(b\) gives \(\log_b(AB)=p+q\), which is \(\log_b A+\log_b B\). This proves the product rule in this setting.
For a difference, \(A/B=b^p/b^q=b^{p-q}\), since \(B>0\). Taking \(\log_b\) gives \(\log_b(A/B)=p-q=\log_b A-\log_b B\). Thus both rules follow from the laws of exponents and the inverse relationship between exponentials and logarithms. The positivity assumptions ensure all the logarithms used in the argument are defined.
A Reliable Solving Process
For an equation made up of sums or differences of logarithms, use this sequence:
Each original logarithm's argument must be positive. These conditions identify values that cannot be solutions.
Use the product rule for a sum and the quotient rule for a difference. Keep the common base unchanged.
For equal logarithms with the same valid base, set their positive arguments equal. For a logarithm equal to a number, convert to exponential form.
Test the candidates in the original equation, not just in the transformed algebraic equation.
When recording positivity conditions, solve only what is needed to determine whether a candidate is allowed. A later tutorial treats domain checks in logarithmic equations more fully; here the essential point is that an invalid logarithm cannot be part of a solution.
Worked Examples
Worked Example: A Sum of Two Logarithms
Solve \(\log_2(x)+\log_2(x-2)=3\).
First, the arguments must be positive. The conditions are \(x>0\) and \(x-2>0\), so \(x>2\).
Both logarithms have base \(2\), so combine their sum with the product rule:
The equation becomes \(\log_2\bigl(x(x-2)\bigr)=3\). Convert to exponential form, then expand:
Factor the quadratic and solve:
The condition \(x>2\) rejects \(x=-2\). To check \(x=4\) in the original equation, the two arguments are \(4\) and \(4-2=2\). Thus:
The left side equals the right side, and both logarithms are defined. Therefore the solution is \(x=4\).
Worked Example: A Difference of Logarithms
Solve \(\log_2(x+2)-\log_2(x-2)=2\).
The arguments must satisfy \(x+2>0\) and \(x-2>0\). Together these require \(x>2\).
Use the quotient rule, preserving the order of the terms in the difference:
Convert the equation to exponential form:
Because \(x>2\), the denominator is nonzero. Multiply both sides by \(x-2\) and solve:
Check the candidate in the original equation. Its arguments are \(\frac{10}{3}+2=\frac{16}{3}\) and \(\frac{10}{3}-2=\frac{4}{3}\), both positive. Their ratio is:
Therefore the left side of the original equation is \(\log_2 4=2\), as required. The solution is \(x=\frac{10}{3}\).
Worked Example: Two Candidates, Only One Valid
Solve \(\log_4(x)+\log_4(x-3)=1\).
The argument conditions are \(x>0\) and \(x-3>0\), so \(x>3\). Combine the logarithms using the product rule:
Convert the resulting equation to exponential form and solve the quadratic:
Since \(x>3\), the candidate \(x=-1\) is invalid. For \(x=4\), the original logarithms are defined, and direct substitution gives:
The check confirms that \(x=4\) solves the original equation. It is the only solution.
Common Pitfall: Adding the Arguments
A frequent error is to treat a sum of logarithms as the logarithm of a sum. The product rule says to multiply the arguments, not add them. For example, the following calculation is wrong:
To see the problem concretely, let \(x=3\). The original sum is \(\log_2(4)+\log_2(2)=2+1=3\). The incorrect expression would be \(\log_2(4+2)=\log_2(6)\), which is not \(3\), because \(2^3=8\), not \(6\). The correct combination is:
For differences, another error is reversing the quotient. The order must match the original subtraction: \(\log_b A-\log_b B=\log_b(A/B)\), not \(\log_b(B/A)\). A useful check is to write the numerator from the first logarithm and the denominator from the second.
What the Algebra Does—and Does Not—Guarantee
Combining logarithms and solving the resulting equation preserves the intended equation only when the logarithms involved are defined and the log laws are used correctly. But algebraic transformations can produce roots that fail the original argument conditions. A quadratic, for instance, may have two roots even when only one makes every logarithm defined.
For that reason, keep the original equation visible while solving. After finding candidates, check two things: every original logarithm has a positive argument, and substitution makes the original left and right sides equal. A candidate that fails either test is not a solution. Never accept a value solely because it solves the polynomial or rational equation obtained during the work.
Check Your Understanding
For each question, pay attention to both the log law and the candidate check.
- Which log law combines \(\log_5 A+\log_5 B\), and what conditions on \(A\), \(B\), and the base are required?
- Solve \(\log_3(x)+\log_3(x-2)=2\). Which candidate satisfies the original equation?
- Combine \(\log_7(x+4)-\log_7(x)\) as a single logarithm. State the conditions needed for the combination.
- Explain why solving the transformed algebraic equation does not, by itself, prove that every resulting candidate solves the original logarithmic equation.
- A learner replaces \(\log_2(x+3)+\log_2(x-1)\) with \(\log_2(2x+2)\). Identify the mistake and give the correct combined expression.
