Logarithmic equations: which solutions survive?

Logarithmic equations begin with the domain

For a real logarithm log_b(u), the base must satisfy b > 0 and b ≠ 1, while the argument must satisfy u > 0. The base is usually a fixed valid number in school exercises, so the variable restriction comes from each argument. In log(x) + log(x − 3) = 1, both x > 0 and x − 3 > 0 are required. Their intersection is x > 3. Write this restriction before combining or solving anything.

Every logarithm gets its own inequality. For ln[(x − 1)/(x + 2)], the whole quotient—not numerator and denominator separately—must be positive, and x cannot equal −2. A sign chart is safer than requiring both parts to be positive, because a quotient is also positive when both parts are negative. The optical aperture in the photograph is an analogy for an allowed opening, not a mathematical graph. The exact domain comes from inequalities, never from visual proportions.

• Review logarithmic equations and extraneous solutions in OpenStax

Convert one logarithm to exponential form

The definition log_b(u) = c means exactly b^c = u. For log₂(x − 1) = 3, convert to 2³ = x − 1, so x = 9. The domain requires x − 1 > 0, or x > 1, and 9 satisfies it. Substitution gives log₂(8) = 3. Notice that the logarithm does not disappear by an arbitrary cancellation; exponential form is an equivalent statement only when the base and argument conditions hold.

Keep the base attached to the logarithm. log₂(x) = 3 and log₁₀(x) = 3 have solutions 8 and 1,000 respectively. A small subscript lost in a photo changes the problem drastically. Natural logarithm ln uses base e, while an unmarked log commonly means base ten in many algebra courses, but conventions vary. Read the course notation and copy it explicitly rather than guessing from a calculator button.

Capture the complete equation before requesting help

A useful photo must show the entire equation, every logarithm base, all parentheses, equality sign, and any instruction about exact or decimal form. Include the line above if it defines a variable or base. Remove names, student numbers, faces, and unrelated personal details, but do not crop away a minus sign or closing bracket. The observatory photograph contains only abstract circles because generated images should not pretend to display a real solver interface or a reliable mathematical answer.

Before reading Eqora's explanation, compare its recognized equation character by character with the original. Is x − 3 still inside the second logarithm? Is the right side 1 rather than 7? Are both logs base ten? Retake a blurred or angled photo. Then state your own domain first. This prevents a correct solution to a misread problem and gives you a concrete checkpoint against which to inspect the generated steps.

• Photograph a complete math problem clearly

Combine logarithms only when the rules apply

For the same base, log_b(M) + log_b(N) = log_b(MN), and log_b(M) − log_b(N) = log_b(M/N), provided the original arguments are positive. A coefficient can become an exponent: k log_b(M) = log_b(M^k). These rules do not say log(M + N) = log M + log N; that tempting identity is false. Test M = N = 1: log(2) is not log(1) + log(1) = 0.

In the opening example, combining is valid only on x > 3. Then log[x(x − 3)] = 1 converts to x(x − 3) = 10. Solving the quadratic creates candidates, not guaranteed answers. The polynomial equation has a larger domain than the original logarithmic equation, which is why −2 appears. Preserve the restriction beside every line rather than hoping to remember it after factoring.

Inspect the first rejected candidate and ask why

When Eqora shows x = 5 and x = −2 before keeping only 5, inspect that transition. Ask a focused question such as, ‘Which original logarithm makes −2 invalid?’ The answer should refer to the original arguments: x = −2 gives −2 and −5, neither positive. Do not settle for ‘negative answers are not allowed.’ A variable may be negative while its logarithm argument remains positive, as happens with ln(x²) when x = −1.

Explain the rejection in your own words and substitute the surviving value. For x = 5, log(5) + log(2) = log(10) = 1. If one line is unclear, ask about that line rather than requesting the whole answer again. The two observers and nested filters represent examining one transition; the blank cards contain no calculations. A precise question helps you repair the method instead of copying a conclusion.

Equal logarithms let you equate positive arguments

If log_b(S) = log_b(T) with the same valid base and both arguments positive, then S = T. For ln(2x − 1) = ln(x + 5), the domain is 2x − 1 > 0 and x + 5 > 0, hence x > 1/2. Equating arguments gives 2x − 1 = x + 5, so x = 6. Both arguments equal 11, confirming the solution.

A candidate can be negative and still work. OpenStax's example ln(x²) = ln(2x + 3) leads to x² = 2x + 3 and candidates x = 3 and x = −1. The domain requires x ≠ 0 and x > −3/2; both candidates satisfy it. Substitution confirms both sides are ln(9) for 3 and ln(1) for −1. Check arguments, not the sign of x in isolation.

• Review the quadratic step that produces candidates

Two logarithms can create a quadratic candidate set

Solve log₃(x − 1) + log₃(x − 3) = 2. The domain is x > 3. Combine to log₃[(x − 1)(x − 3)] = 2, then convert: (x − 1)(x − 3) = 9. Expanding gives x² − 4x − 6 = 0, so x = 2 ± √10. Since 2 − √10 is below 3, it is invalid. The solution is x = 2 + √10.

Check without rounding first. For x = 2 + √10, the arguments are 1 + √10 and √10 − 1, both positive. Their product is (√10 + 1)(√10 − 1) = 9, so the two logs combine to log₃(9) = 2. This exact check is stronger than substituting a prematurely rounded decimal. It confirms both the domain and the transformed equation.

Close the explanation and solve a transfer problem

Now hide the worked steps and solve log₂(x) + log₂(x − 2) = 3. Write the domain x > 2. Combine to log₂[x(x − 2)] = 3, convert to x(x − 2) = 8, and solve x² − 2x − 8 = 0. The candidates are 4 and −2. Only 4 belongs to x > 2, and substitution gives log₂(4) + log₂(2) = 2 + 1 = 3.

Use Eqora only after completing that transfer attempt. Compare the domain, transformation, candidates, and final substitution in that order. Ask about the first disagreement, then close the explanation and redo the calculation. Follow your school or university rules about permitted tools. Eqora can misread a base or make a reasoning error; it cannot replace your responsibility to verify the original notation, domain, algebra, exact form, and final answer.

• Check equation candidates independently

Put it into practice now

Choose one problem from your current homework or review set. Attempt it before opening Eqora, then use the app only at the point where your own reasoning stops.

• State what the problem is asking before you solve it
• Identify the first step you cannot justify
• Ask Eqora one focused follow-up about that step
• Finish with a similar problem and no solution in view

The session is complete when the method is clearer, not simply when the worksheet has one more answer.