Did you know? The formula ln(x²) = 2ln x looks completely familiar—and is incomplete if x can be negative. For x = −3, ln(x²) = ln 9 is a perfectly valid real number, while ln(−3) is not a real logarithm. The accurate real-variable identity is ln(x²) = 2ln|x|, for x ≠ 0.
For JC1 H2 Mathematics learners in Bukit Timah, that one pair of absolute-value bars can protect the domain of an entire solution. The 2027 Singapore–Cambridge A-Level H2 Mathematics syllabus 9758 assumes logarithmic laws and functions from O-Level Additional Mathematics, and develops further calculus involving logarithmic expressions. It is common to know how to rearrange indices but forget that a logarithm’s argument must be strictly positive.
This guide explains why ln(x²) is different from 2ln x, why absolute values appear in differentiating and integrating, and why solving a logarithmic equation requires checking the original domain. You will find exact counterexamples, worked algebraic cases, derivative checks, careful interval reasoning and short practice questions. The goal is a reliable method, not a collection of rules repeated without meaning.
Quick answer: square first, then take the logarithm
For real x ≠ 0, x² is strictly positive. Therefore ln(x²) is defined for all negative and positive values of x, but not for x = 0.
The expression 2ln x is defined only for x > 0, because the real natural logarithm ln x requires a positive input.
So the identities must be stated with their domains:
- ln(x²) = 2ln|x| for all real x ≠ 0.
- ln(x²) = 2ln x is valid for x > 0.
- ln(x²) = 2ln(−x) is valid for x < 0.
- ln|x| is defined for x ≠ 0 and has derivative 1/x on that domain.
- ln(x² + 1) is defined for every real x because x² + 1 ≥ 1.
- ln[(x − a)²] is defined for x ≠ a and equals 2ln|x − a| on that domain.
The central habit is simple: work out where the original logarithm exists before applying an algebraic identity.
What a logarithm is allowed to receive
The real natural logarithm ln t is defined for a strictly positive real argument t. It is the inverse function of eˣ, which produces positive outputs for every real input.
That explains the input restriction. There is no real number y for which eʸ = −3, so ln(−3) is not defined as a real-valued function. In a complex-number setting, logarithms can be extended differently, but that is not what ordinary real-variable H2 algebra questions mean.
Now let t = x². The input to the logarithm is not x, but x². This input is positive whenever x is non-zero. A negative x is perfectly acceptable because its square is positive.
For x = −3, the input x² is nine; for x = +3, the input is also nine. Hence ln(x²) gives the same value at these two x-values. That makes the function even.
At x = 0, however, the input is zero. Since ln 0 is not a finite real number, the function has a missing input there. We must not include it just because zero is allowed for the polynomial x² by itself.
Why the ordinary power law has conditions
Students are often taught a law such as ln(aⁿ) = n ln a. In its familiar real-valued form, the base a is positive. With a > 0, the law is valid for real powers where the expressions are defined.
If we set a = x when x is negative, we have violated that positive-base requirement. We cannot automatically write ln(x²) = 2ln x because ln x is not a real number there.
The correct way to handle a square of a non-zero real x is to notice that x² = |x|² and |x| is positive. Then the logarithm law applies:
ln(x²) = ln(|x|²) = 2ln|x|.
The absolute value restores a positive base, and the resulting expression has the same domain as the original.
This is a powerful pattern: first rewrite an expression in a form where the law’s assumptions hold, then apply the law. Do not apply it outside its domain and hope the algebra becomes true afterward.
Worked example 1: test a positive and a negative input
Let x = 3. The original expression gives ln(3²) = ln 9. The expression 2ln 3 also gives ln 9, and 2ln|3| is the same. All three work because x is positive.
Now let x = −3. The original expression gives ln[(-3)²] = ln 9. The expression 2ln|−3| gives 2ln 3 = ln 9, so it still works.
But the expression 2ln(−3) is not a real number. It cannot equal ln 9 in ordinary real-variable algebra. That single substitution disproves the unqualified identity.
A useful classroom check is to ask a learner to test one positive and one negative value whenever a logarithm law is applied to a variable expression. It reveals domain problems quickly without requiring a long proof.
Worked example 2: solve ln(x²) = ln 9
We want real x satisfying ln(x²) = ln 9. Both logarithm arguments are positive on the permitted domain x ≠ 0.
The natural logarithm is one-to-one on positive real numbers, so equality of the logarithms implies equality of their arguments:
x² = 9.
Hence x = 3 or x = −3.
Check both in the original equation. At x = 3 and x = −3, x² = 9 and the left side equals ln 9. Both are valid.
If we had rewritten ln(x²) as 2ln x and then solved, we would have considered only x > 0 and lost the valid negative solution. The algebra would look shorter but the solution set would be incomplete.
The error is about lost domain, not a difficult quadratic
The equation x² = 9 is elementary. A learner who gives only x = 3 is not necessarily struggling with solving quadratics; they may have silently changed the original domain by choosing an invalid logarithm identity.
The correct diagnostic is to ask: “Which values of x made the original logarithm argument positive?” Both positive and negative non-zero x qualify. This restores the missing branch.
That approach generalises. Whenever a transformation of an equation narrows the domain, ask whether valid solutions may have been discarded.
Worked example 3: a shifted square
Solve ln[(x − 1)²] = ln 16.
The original expression is defined for x ≠ 1. Equality of the positive arguments gives (x − 1)² = 16.
Therefore x − 1 = 4 or x − 1 = −4. Solving yields x = 5 or x = −3.
Both satisfy the original domain restriction and produce argument sixteen. The negative solution is not a mistake. The square of x − 1 can be sixteen on either side of the centre x = 1.
The matching logarithm identity is ln[(x − 1)²] = 2ln|x − 1| for x ≠ 1. If we incorrectly wrote 2ln(x − 1), the formula would only exist for x > 1 and would discard x = −3.
A small horizontal shift changes the excluded input from zero to one, but the same logical structure remains.
The graph of ln(x²): two branches, one asymptote
The function f(x) = ln(x²) has domain x < 0 or x > 0. It is symmetric about the y-axis because replacing x by −x does not change x².
As x approaches zero from either side, x² approaches zero through positive values, so ln(x²) decreases without bound. The vertical asymptote is x = 0.
At x = −1 and x = 1, the function equals ln 1 = 0. Thus it crosses the x-axis at both −1 and 1. For |x| > 1, x² > 1 and the logarithm is positive. For 0 < |x| < 1, the logarithm is negative.
As |x| becomes large, ln(x²) increases without bound. The left and right branches mirror each other but remain separated by the excluded input x = 0.
A sketch with only the right branch has effectively drawn 2ln x rather than the complete ln(x²) function. Checking symmetry is an efficient way to catch the omission.
Differentiate ln(x²) without losing its domain
Let f(x) = ln(x²) for x ≠ 0. Use the chain rule: the derivative of ln u is u′/u when u is positive.
Here u = x² and u′ = 2x. Therefore
f′(x) = (2x)/(x²) = 2/x, for x ≠ 0.
This formula is valid on both negative and positive parts of the original domain. For example, at x = −2, the derivative is −1. The left branch is decreasing as x increases towards zero.
Differentiate the alternative expression 2ln|x|. The derivative of ln|x| is 1/x for x ≠ 0, so its derivative is again 2/x. The two descriptions agree.
Now compare with 2ln x. Its derivative is also 2/x, but its original domain is only x > 0. Matching derivative formulas does not imply that the original functions have identical domains.
Why ln|x| appears in integration
Consider the integral of 2/x with respect to x, on any interval that does not contain zero.
An antiderivative is 2ln|x| + C, since its derivative is 2/x for x ≠ 0.
We could also write ln(x²) + C on such an interval, because ln(x²) = 2ln|x| wherever x is non-zero.
Writing 2ln x + C would only describe an antiderivative on intervals with x > 0. It would not be a real-valued expression for the interval x < 0.
There is a further subtlety: the domain x ≠ 0 is disconnected into two intervals. An antiderivative can have different additive constants on x < 0 and x > 0 when those intervals are considered separately. In routine exercises, writing +C often assumes we are working on a single connected interval.
This is the reason the familiar integration formula ∫(1/x)dx = ln|x| + C contains an absolute value.
Worked example 4: a product inside a logarithm
Consider f(x) = ln[(x + 1)(x − 4)]. First determine where the argument is positive.
The factors are zero at x = −1 and x = 4. Their product is positive for x < −1, when both factors are negative, and for x > 4, when both are positive. Thus the domain is
x < −1 or x > 4.
On this domain, we may write
ln[(x + 1)(x − 4)] = ln|x + 1| + ln|x − 4|.
Both absolute-value arguments are positive, so the right-hand logarithms are defined on exactly the intervals we need. Notice that the expression ln(x + 1) + ln(x − 4) would fail on x < −1 because both individual arguments there are negative.
Differentiate on the permitted intervals:
f′(x) = 1/(x + 1) + 1/(x − 4).
We can check directly by the chain rule. The product inside the logarithm is x² − 3x − 4, whose derivative is 2x − 3. Therefore f′(x) = (2x − 3)/(x² − 3x − 4).
Combine the two fractions on the other route and obtain the same expression. The derivative checks agree without sacrificing the domain.
Worked example 5: a logarithmic equation with an invalid algebraic root
Solve ln(x + 2) + ln(x − 3) = ln 4 for real x.
Before combining, the two logarithms require x + 2 > 0 and x − 3 > 0. Together they demand x > 3.
Now the product law applies because both arguments are positive. Combine:
ln[(x + 2)(x − 3)] = ln 4.
Hence (x + 2)(x − 3) = 4. Expand:
x² − x − 6 = 4, so x² − x − 10 = 0.
The quadratic formula gives x = [1 ± √41]/2.
The positive branch [1 + √41]/2 is approximately 3.70156, which satisfies x > 3. The negative branch [1 − √41]/2 is approximately −2.70156, which violates the original logarithm domain and must be rejected.
Thus the only valid real solution is x = (1 + √41)/2.
Why the rejected root was not a calculation error
The quadratic equation was derived correctly after combining the logarithms, but it has a real solution that is not an allowed input for the original equation. The original requirement x > 3 is decisive.
If a student merely copies both quadratic roots as the final answer, they have confused solutions of a transformed algebraic equation with solutions of the starting logarithmic equation.
The repair is to write down the original domain before any combination and check every candidate afterward. That simple habit also applies to rational equations, square-root equations and inverse trigonometric expressions.
What if the logarithm contains x² + 1?
Let g(x) = ln(x² + 1). Since x² + 1 ≥ 1 for every real x, its logarithm is defined on the whole real line.
Differentiate using the chain rule:
g′(x) = 2x/(x² + 1).
This is not equal to 2ln|x|, and ln(x² + 1) cannot be split as ln(x²) + ln 1. The logarithm law does not turn logarithms of sums into sums of logarithms.
A basic numerical check disproves any such claim. At x = 1, ln(1² + 1) = ln 2, whereas ln(1²) + ln 1 = 0 + 0 = 0. The expressions differ.
The valid product law concerns multiplication, not addition. This is another example where identifying the outermost algebraic operation protects the solution.
A quick decision method for any real logarithm
When a logarithm appears, ask four questions in order:
- What is the entire argument? Include all brackets and factors.
- Where is that argument strictly positive? Solve this before simplifying.
- Which logarithm law is being used? Check that its required positive inputs are actually permitted.
- Does the final expression retain the correct domain? Reject candidate solutions that do not satisfy the original condition.
This routine works for ln(x²), ln[(x − 3)²], ln(x² + 1), and products or quotients inside the logarithm.
You do not need to remember a new trick for every chapter. The condition “argument > 0” is the stable starting point.
A three-minute diagnostic for students
First ask whether ln(9) and ln(−9) are both real. The learner should know that only ln(9) is a real logarithm.
Then ask why ln[(-3)²] exists despite x = −3 being negative. The answer should focus on the squared argument, not on a claim that negative logarithms suddenly became allowed.
Third, present ln(x²) and 2ln x and ask for a negative test input. A learner who chooses x = −2 or x = −3 and explains the domain mismatch has grasped the issue.
Finally ask for the derivative of ln(x²) and its domain. The correct answer is 2/x for x ≠ 0. This connects algebraic domains to calculus rather than treating them as separate chapters.
Practice: ten questions with answers
Question 1. State the real domain of ln(x²). Answer: x ≠ 0.
Question 2. Simplify ln(x²) using a valid real identity. Answer: 2ln|x| for x ≠ 0.
Question 3. Is ln[(−4)²] defined as a real logarithm? Answer: Yes; its argument is sixteen.
Question 4. Is 2ln(−4) a real number? Answer: No; ln(−4) is undefined over the reals.
Question 5. Solve ln(x²) = ln 25. Answer: x = −5 or x = 5.
Question 6. Solve ln[(x + 2)²] = ln 9. Answer: (x + 2)² = 9, so x + 2 = ±3 and x = 1 or x = −5.
Question 7. Differentiate ln(x²) on its real domain. Answer: 2/x for x ≠ 0.
Question 8. Differentiate ln(x² + 1). Answer: 2x/(x² + 1), valid for all real x.
Question 9. State the real domain of ln[(x − 1)(x + 3)]. Answer: x < −3 or x > 1.
Question 10. Evaluate ∫(1/x)dx on an interval avoiding zero. Answer: ln|x| + C.
To test independent understanding, ask a learner to justify questions one, five and nine. The answers depend on a consistent domain principle, not on memorising an extra logarithm law.
The difference between an algebra mistake and a domain mistake
Suppose a student writes ln[(x − 1)²] = 2ln(x − 1). For x > 1, the expression is valid. For x < 1, it is not a real identity because the right side is undefined. The algebraic shape looks reasonable; the declared domain is too narrow.
Now suppose the same student writes ln(x² + 1) = ln(x²) + ln 1. This is a different mistake: the logarithm law was applied to a sum instead of a product. The expression is wrong even for many positive x-values.
A tutor should distinguish these causes. The first needs attention to positive logarithm bases and absolute values. The second needs attention to the exact operations covered by logarithm laws.
The repair is specific, concise and testable. Once the learner understands the rule, an unseen question can confirm that the habit transfers.
A focused three-stage JC1 tutorial route
Stage one: positive logarithm arguments
Work through ln x, ln(x²), ln(x² + 1) and ln[(x − 1)²]. Ask for domains before values. Show a simple graph when helpful.
Stage two: valid identities and derivative checks
Use ln(x²) = 2ln|x| for x ≠ 0, then differentiate both sides. Next compare with ln(x² + 1) and explain why it cannot be split using a sum law.
Stage three: equations with candidate-root checks
Give logarithmic equations that produce two algebraic candidates but only one permitted input. Require the student to write the domain at the beginning and substitute candidate roots back into the original equation.
A strong final check is whether the student can explain why a negative x can be allowed in ln(x²) but not in ln x, without simply repeating the phrase “absolute value”.
Frequently asked questions
Is ln(x²) the same as 2ln x?
They agree for x > 0. The correct real identity on the full domain x ≠ 0 is ln(x²) = 2ln|x|.
Why can x be negative in ln(x²)?
Because the input to ln is the square x², which is positive for every non-zero real x.
Is ln|x| defined at zero?
No. |0| = 0, and the real natural logarithm is not defined for input zero.
Why does the derivative of ln(x²) equal 2/x?
By the chain rule, d[ln(x²)]/dx = (2x)/x² = 2/x for x ≠ 0. This agrees with d[2ln|x|]/dx.
Can I split ln(a + b) as ln a + ln b?
No. The logarithm product law is ln(ab) = ln a + ln b for positive a and b. It does not apply to sums.
Does a logarithmic equation always accept every quadratic root obtained after simplifying?
No. Algebraic transformations can produce candidates outside the original logarithms’ domains. Check each candidate in the starting equation.
Is this on the H2 syllabus?
Logarithmic functions and laws form part of the assumed O-Level Additional Mathematics knowledge listed in the 2027 H2 Mathematics syllabus 9758; the concepts continue to support H2 differentiation, integration and modelling.
Read further and choose an appropriate tuition route
For another example of a condition that must survive simplification, see why some composite functions do not exist. For calculus structure, read why the product rule has two terms. The official 2027 H2 Mathematics syllabus is the source for examined topics and the prerequisites students are assumed to know.
The immutable eduKateSG Secondary 1 Mathematics tutorial reference offers a model of diagnosing a precise misunderstanding, practising and checking independence. It does not set JC1 pricing, group size or class availability. For current JC1 arrangements, visit Bukit Timah JC1 tuition.
One question should accompany every logarithm: is the entire argument positive? That is the key that keeps both the mathematics and its domain intact.