Condense to a single logarithm, convert to exponential form, solve, and then test every candidate in the original equation. That last step is not diligence, it is arithmetic: condensing enlarges the set of values the expression accepts, so it can manufacture solutions the original never had.
Core ideas
Two moves solve every equation in this lesson, and one property of the first move explains every wrong answer.
There are two moves. Condense several logarithms into one with the laws of “The Laws of Logarithms”, then convert log_b(u) = c into u = b^c by the definition in “What a Logarithm Is”. Nothing is cancelled and nothing is divided by; the definition is quoted, in the direction that removes the logarithm.
log M + log N requires M > 0 and N > 0. log(MN) requires only MN > 0, which two negatives also satisfy. The two agree wherever both are defined, and the second is defined in more places. Every extraneous solution in this lesson enters through exactly that gap.
A candidate is rejected when it makes some argument of some logarithm zero or negative. x = −3/2 solves log_3(x + 6) − log_3(x + 2) = 2 perfectly well, because at that value the arguments are 9/2 and 1/2. Rejecting negative candidates on sight throws away correct answers.
The material
Two solving moves, the precise reason a false answer can appear, and the three ways the solution set can come out empty.
In “Solving Exponential Equations” the unknown sat in an exponent and a logarithm brought it down. Here the unknown sits inside a logarithm, and an exponential takes it out. The two lessons are the same manoeuvre in opposite directions, with one asymmetry: an exponential accepts every real number, and a logarithm does not.
An equation in which the unknown appears in the argument or the base of a logarithm. Its domain is the set of values making every argument positive (and every base positive and not 1). The domain is fixed by the equation as given, and no later step may enlarge it.
For b > 0, b ≠ 1: log_b(u) = c if and only if u = b^c, for u > 0 and any real c.
And log_b(M) = log_b(N) if and only if M = N, for M, N > 0. The second is the first applied twice, and it is the statement that lets two matching logarithms be dropped from an equation.
The method follows from the two conversions. Condense until each side is a single logarithm or a constant. If both sides are logarithms in the same base, drop them. If one side is a constant, convert. Then solve whatever is left, which is usually linear or quadratic.
Look at why x = −3 appeared at all. The condensed equation asks only that (x − 1)(x + 2) equal 4, and at x = −3 those factors are −4 and −1, whose product is 4. Two negatives passed a test that was written for their product. The original equation asks for the logarithm of each factor separately, and neither exists.
log_3(x + 6) − log_3(x + 2) = 2 condenses to log_3((x + 6)/(x + 2)) = 2, so (x + 6)/(x + 2) = 9, giving x + 6 = 9x + 18 and x = −3/2. The arguments there are 9/2 and 1/2, both positive, so the candidate stands. Check: log_3(9/2) − log_3(1/2) = log_3(9) = 2.
log_2(x) − log_2(x + 1) = 3 condenses to x/(x + 1) = 2^3 = 8, so x = 8x + 8 and x = −8/7. That value makes both arguments negative, so it is rejected and nothing else is available. The solution set is empty, and no step along the way was wrong.
A third ending is possible: the condensed equation can be a contradiction. log(3x + 1) = log x + log 3 becomes log(3x + 1) = log(3x), hence 3x + 1 = 3x, hence 1 = 0. No candidate is ever produced, and the solution set is empty for a reason that has nothing to do with domains.
Substitute into the equation you were given, never into the condensed form. The condensed form is precisely the form that accepts extraneous values, so it will confirm them cheerfully. Checking there proves nothing at all.
An equation may also carry the unknown in the base. log_x(16) = 2 converts to x² = 16, so x = 4 or x = −4. A base must be positive and not 1, so −4 is rejected and x = 4 is the answer. The same discipline applies: convert by the definition, then test each candidate against the conditions the definition imposes.
Interactive
The domain check is a step, not a formality
Method
Four steps. The first one is written before any algebra and used after all of it.
Every argument must be positive; intersect those conditions. Do it now, because the condensed equation has forgotten what the conditions were and cannot be asked.
Use the three laws of “The Laws of Logarithms” from right to left. A coefficient goes back up as a power first: 2 log x becomes log(x²), and only then can it join a sum.
Two logarithms in the same base: drop both and equate the arguments. A logarithm equal to a constant: apply the definition, u = b^c. What is left is an ordinary equation with no logarithms in it.
Keep the candidates inside the domain and discard the rest, naming the argument that went non-positive. If nothing survives, the solution set is empty and that is the answer.
The domain line costs one line at the top and does all the deciding at the bottom.
The rejected candidate is not a mistake in the algebra. At x = −2 the two factors are −2 and −4, and their product is 8 exactly as the condensed equation demands. It satisfies every line below the condensing step and none of the equation above it.
Note what the check is testing. Not "is −2 negative" but "are the arguments x and x − 2 positive at x = −2". Those are different questions, and only the second one is the domain condition: the reading's x = −3/2 is a negative candidate that passes it and is a genuine solution.
For a faded domain-and-solve pass, use log_2(x − 1) + log_2(x − 3) = 3. The supplied first line is the original domain x > 3. Condense, convert, factor, and test both candidates in the original two arguments. Your final line must name which argument rejects any discarded candidate, not merely call the candidate negative.
Given log(x − 3) + log x = 1, you should write x > 3 before condensing anything, and then be able to say in advance that one of the two candidates from x² − 3x − 10 = 0 will fail it. It does: x = 5 is the solution and x = −2 is rejected.
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