How do I solve a logarithmic equation and check for invalid answers?

Write down the domain restrictions first: each expression inside a logarithm must be strictly positive. A logarithm of zero or a negative number is not defined in the real numbers, so any value that makes an input nonpositive cannot be a real solution. [S1]

For example, in ln(x−1)+ln(x+1)=ln(2x), the three inputs are x−1, x+1, and 2x. They must all be positive. Together, those conditions require x>1. Keep that restriction visible while doing the algebra; it will help you identify a candidate that must be discarded.

Start with the equation:

ln(x−1)+ln(x+1)=ln(2x)

For positive inputs, the sum of two natural logarithms can be combined as the logarithm of their product. The domain conditions are already recorded, so combine the left side and equate the positive arguments:

1. ln((x−1)(x+1))=ln(2x)
2. (x−1)(x+1)=2x
3. x²−1=2x
4. x²−2x−1=0
5. x=1±√2

The domain restriction x>1 still applies. The algebra produces two candidates, but it does not automatically make both valid. Applying the domain now, 1−√2 is less than 1, while 1+√2 is greater than 1.

Algebraic steps can produce candidates outside the original equation’s domain. The original logarithms are defined only when their inputs are positive, so a candidate that violates that condition is not a solution. Checking candidates in the original equation is the reliable final step for catching such extraneous answers. [S1]

Here, x=1−√2 fails x>1, so discard it. For x=1+√2, the inputs are x−1=√2, x+1=2+√2, and 2x=2+2√2; all are positive. Also, (x−1)(x+1)=x²−1=2x, so the two sides of the original logarithmic equation have matching positive arguments. Thus the valid solution is x=1+√2. The domain check and substitution verify this example; the algebraic candidates alone would not.

Use this copyable routine:

1. List inputs: Set each logarithm’s argument greater than zero and find the values satisfying all conditions.
2. Solve: Apply logarithm rules only when their inputs meet those conditions. Keep track of any equation changes.
3. Find candidates: Solve the resulting equation, but label the values as candidates rather than solutions.
4. Filter by domain: Discard any candidate that makes even one original logarithm’s argument zero or negative.
5. Substitute: Put each remaining candidate into the original equation and confirm both sides are defined and equal.

If an equation has a complicated domain or several algebraic branches, show those steps separately rather than skipping directly from candidates to an answer. The key is that a value must satisfy both the original equation and its domain.