Why Domain Checks Matter
A logarithmic equation is defined only when every logarithm in it has a positive argument. An algebraic transformation can sometimes produce values that solve a transformed equation but make a logarithm in the original equation undefined. Such values are called extraneous candidates.
The reliable method is to identify the domain first, solve the equation, and then check each candidate in the original equation. The domain is not an optional final detail: it is part of the problem from the beginning.
Finding the Domain of a Logarithmic Equation
To find the domain, list the argument of each logarithm and require it to be positive. The allowable values of the variable must satisfy all those conditions at once. The domain is their intersection.
For example, in \(\log(x-1)+\log(x+2)\), the two arguments are \(x-1\) and \(x+2\). They must both be positive, so \(x>1\) and \(x>-2\). A value must satisfy both conditions, which leaves \(x>1\).
This check should be made before using a logarithm identity. The product rule, for instance, says \(\log_b(u)+\log_b(v)=\log_b(uv)\) when \(u>0\) and \(v>0\). If either original argument is not positive, the original sum of logarithms is not defined, even if the product \(uv\) happens to be positive.
Equations with Matching Logarithms
When the bases are the same, logarithms are one-to-one on their domains. Thus, for \(b>0\), \(b\ne1\), and positive \(u\) and \(v\), the equation \(\log_b(u)=\log_b(v)\) is equivalent to \(u=v\). The positivity conditions remain essential: equating arguments does not make an undefined logarithm valid.
Worked Example: Matching Logarithms with a Domain Restriction
Solve \(\log_3(x+1)=\log_3(5-x)\).
First require both arguments to be positive:
The domain is therefore \(-1<x<5\). Since the bases match and both arguments are positive on this domain, equate the arguments:
Add \(x\) to both sides and subtract \(1\) from both sides:
The candidate \(2\) is in the domain. Check it in the original equation: \(x+1=3\) and \(5-x=3\), so both sides are \(\log_3(3)=1\). The solution is \(x=2\).
Why Transformations Can Create Extraneous Candidates
A transformation is safe only when it preserves the meaning of the equation for the values under consideration. Logarithm identities have domain conditions, and algebraic equations formed after applying those identities may have solutions outside the original domain.
For example, combining two logarithms into one replaces their separate arguments with a product. A positive product does not guarantee that each factor is positive. The combined expression may therefore be defined at values where the original logarithms are not. Those values must not be accepted as solutions.
Here is the reasoning. A solution of the original equation must first make every original logarithm defined, so it belongs to \(D\). If the steps used to obtain \(P(x)=0\) are valid on \(D\), that solution must also satisfy the resulting equation. This establishes why solving the algebraic equation gives candidates rather than an automatic answer: its roots may include values outside \(D\), and a transformation that is not reversible can require additional checking.
Combining Logarithms and Filtering Candidates
When adding logarithms with the same base, first record the positivity conditions on their separate arguments. Then combine them, solve the resulting equation, and retain only candidates satisfying the original conditions. Finally substitute the retained candidates into the original equation.
Worked Example: A Candidate Rejected by the Domain
Solve \(\log_2(x-1)+\log_2(x+1)=1\).
The separate arguments must both be positive:
Both conditions together give the domain \(x>1\). On this domain, the product rule applies:
Write the right side as a logarithm, \(1=\log_2(2)\). Both arguments are positive on the original domain, so matching the logarithms gives
Expand and solve:
The domain requires \(x>1\), so \(-\sqrt{3}\) is rejected and \(\sqrt{3}\) remains. Verify it in the original equation. Its arguments are positive because \(\sqrt{3}>1\), and
Therefore, by the product rule, the left side of the original equation is \(\log_2(2)=1\), equal to its right side. The solution is \(x=\sqrt{3}\).
When the Equation Has No Valid Solution
Sometimes every root of the equation produced by algebra falls outside the original domain. In that case the original equation has no real solutions. A particularly important warning is that a logarithm argument equal to zero is not allowed: logarithms require strictly positive arguments.
Worked Example: No Root Survives the Domain Check
Solve \(\ln(x^2+x-2)=\ln(x-1)\).
Require both arguments to be positive. The right argument gives \(x-1>0\), so \(x>1\). The left argument factors as \((x+2)(x-1)\); when \(x>1\), both factors are positive, so the left argument is positive as well. Thus the domain is exactly \(x>1\).
Both logarithms have the same base \(e\), and their arguments are positive on the domain. Equate the arguments:
Subtract \(x-1\) from both sides:
Thus the algebraic candidates satisfy
Neither candidate belongs to the domain \(x>1\). At \(x=1\), both arguments are \(0\), and \(\ln(0)\) is undefined. At \(x=-1\), both arguments are \(-2\), and the real logarithm of a negative number is undefined. Therefore, the original equation has no real solutions.
A Complete Check: From Domain to Verification
The reasoning can be organized as a short sequence. This order helps prevent two common errors: losing an original restriction while simplifying, and reporting an algebraic root before checking it.
Write down every expression inside a logarithm in the original equation.
Require each listed argument to be strictly positive, and keep values satisfying all conditions.
Apply logarithm identities only when their argument conditions hold, and solve the resulting equation for candidates.
Reject candidates outside the original domain, then substitute each remaining candidate into the original equation.
Checking a candidate in a transformed equation alone is not enough. The transformed equation may have a wider domain than the original, or a step may not be reversible for every algebraic candidate. The original equation is the final test.
Common Pitfall: Treating a Candidate as a Solution
A frequent mistake is to solve an equation obtained after applying logarithm laws and report every root. For instance, in the second worked example, \(x=-\sqrt{3}\) satisfies the equation \(x^2=3\), but it does not satisfy the original logarithmic equation: \(x-1<0\) and \(x+1<0\), so neither logarithm is defined over the real numbers.
The correction is to keep the original domain alongside the algebra throughout the solution. Then test every root against those restrictions and substitute any survivors into the original equation. The domain conditions are not consequences of the final algebra; they come from the original logarithms and must be preserved.
Check Your Understanding
For each problem, begin by identifying the domain before solving.
- Find the domain of \(\ln(x-3)+\ln(x+1)\).
- For \(\log_4(x+2)=\log_4(6-x)\), what conditions must a solution satisfy before the arguments are equated?
- Why can a root of an equation obtained by combining logarithms fail to solve the original equation?
- In \(\ln(x^2-4)=\ln(x-2)\), what positivity conditions must be checked before matching the arguments?
- After solving a logarithmic equation algebraically, what two checks should be made before reporting a solution?
