Some logarithmic equations are quadratic equations in disguise. If the same logarithmic expression appears squared and to the first power, substitute u for that logarithm, solve the quadratic in u, then convert each u-value back into an equation for x. Check the original logarithm domain before accepting the answers.
For example, (log₂x)² − 3log₂x + 2 = 0 becomes u² − 3u + 2 = 0 when u = log₂x. Thus u = 1 or 2, giving x = 2 or 4. A negative u-value can also be valid, because logarithms may be negative even though their arguments must be positive. This guide shows how to handle that distinction, squared-log notation, shifted arguments, logarithmic inequalities and invalid algebraic shortcuts.
Recognise the repeated logarithmic quantity
The expression (log₂x)² means the square of the number log₂x. If log₂x appears elsewhere in the same equation, the two terms have the structure u² and u.
This suggests the substitution u = log₂x. The resulting equation can often be factorised or solved by the ordinary quadratic formula.
The purpose is to simplify the algebraic structure, not to change the meaning of the variable. The original input x and substituted value u are related by x = 2^u, not by x = u.
Before proceeding, write x > 0 because a real logarithm requires a strictly positive argument. That domain condition must remain attached to the problem even after the logarithm temporarily disappears through substitution.
The first complete worked example
Solve (log₂x)² − 3log₂x + 2 = 0, for real x. Start with the domain x > 0.
Let u = log₂x. The equation becomes u² − 3u + 2 = 0. Factorise it as (u − 1)(u − 2) = 0.
Therefore u = 1 or u = 2. Return to the original logarithmic expressions: log₂x = 1 gives x = 2¹ = 2, while log₂x = 2 gives x = 2² = 4.
Both values satisfy x > 0. The final real solution set is x = 2 or x = 4. This is an exact answer with no need to approximate any logarithm numerically.
Check both answers in the original equation
At x = 2, log₂2 = 1. Substitute into the original equation: 1² − 3(1) + 2 = 0.
At x = 4, log₂4 = 2. Substitution gives 2² − 3(2) + 2 = 4 − 6 + 2 = 0.
These checks confirm both values solve the original equation rather than only the quadratic written in u. A student who reports u = 1 and 2 without converting back has not completed the task.
The original equation concerns x, while the substituted quadratic is a helpful intermediate representation. Every final solution must satisfy the original equation and its logarithm domain.
Why u is allowed to be negative
For positive logarithm base b ≠ 1, the logarithm argument must be positive. But the logarithm output can be negative, zero or positive.
For example, log₂(1/4) = −2 because 2^(−2) = 1/4. The argument 1/4 is positive, so this logarithm is perfectly valid.
Thus when u = log₂x, u ranges over all real numbers as x ranges over all positive real numbers. It would be wrong to impose u ≥ 0 merely because x must be positive.
This distinguishes logarithmic substitutions from substitutions such as u = x² or u = 2^x, which inherit non-negative or positive restrictions on u. The allowed values of the new variable depend on what it represents.
Example with a negative substituted root
Solve (log₂x)² + log₂x − 2 = 0. Write x > 0 and set u = log₂x.
The transformed quadratic is u² + u − 2 = 0. Factorise as (u + 2)(u − 1) = 0, giving u = −2 or u = 1.
For u = −2, solve log₂x = −2 to obtain x = 2^(−2) = 1/4. For u = 1, obtain x = 2.
Both outputs are strictly positive and therefore admissible. The two real solutions are x = 1/4 or 2. Discarding the negative u-root would lose a completely valid logarithmic solution.
The common mistake of rejecting small positive arguments
A logarithm such as log₂x is negative when 0 < x < 1. That does not make it undefined.
For x = 1/4, log₂x = −2. The number inside the logarithm is 1/4, which is positive. The negative result is a property of the exponent needed to produce a fraction from base 2.
The restriction x > 0 excludes zero and negative arguments, not every number smaller than one.
Students who confuse these facts can incorrectly lose half of a quadratic-in-logarithm solution set. Ask them to compare the input domain x > 0 with the possible output values of the logarithm, which extend over the entire real line.
Do not confuse a squared logarithm with the logarithm of a square
The expression (log₂x)² is the square of the logarithm’s value. By contrast, log₂(x²) is the logarithm of x squared.
For x > 0, the power law gives log₂(x²) = 2log₂x. It does not generally equal (log₂x)².
For example, at x = 8, log₂x = 3. The squared logarithm is 9, while log₂(x²) = log₂64 = 6. They are different.
Careful brackets are therefore essential. The substitution u = log₂x turns (log₂x)² into u², while it turns log₂(x²) into 2u under the stated positive-domain condition. Reading one expression as the other changes the entire equation.
A mixed notation example
Solve (log₂x)² − log₂(x²) + 1 = 0 for x > 0.
Set u = log₂x. Because x > 0, the power law gives log₂(x²) = 2u. Thus the equation becomes u² − 2u + 1 = 0.
Factorise as (u − 1)² = 0, yielding u = 1. Therefore log₂x = 1 and x = 2.
This illustrates the importance of applying logarithm laws before solving. An equation may contain both squared logarithms and logarithms of squared arguments, and they must be translated differently into the substituted variable.
Shifted logarithmic arguments create a different x-domain
Consider (log₃(x − 1))² − log₃(x − 1) − 2 = 0. The logarithm requires x − 1 > 0, so x > 1.
Let u = log₃(x − 1). The transformed quadratic is u² − u − 2 = 0, or (u − 2)(u + 1) = 0.
For u = 2, x − 1 = 3² = 9, so x = 10. For u = −1, x − 1 = 3^(−1) = 1/3, so x = 4/3.
Both are greater than 1 and hence valid. The transformed quadratic is ordinary, but returning to x requires remembering that the logarithm’s argument is x − 1, not x.
Recover the original variable in two stages
Whenever u = log_b g(x), solving the quadratic gives possible u-values. Each leads to the equation g(x) = b^u.
The first step is exponential conversion; the second is solving g(x) for x. If g(x) = x − 1, add one. If g(x) = 2x + 3, subtract three and divide by two.
A student who writes x = b^u immediately for every logarithm equation will make errors whenever the argument is not exactly x.
Write the intermediate equation with the complete argument visible. That also supports a final check that the argument remains strictly positive, as real logarithms require.
A quadratic logarithm with the quadratic formula
Solve (log₂x)² − log₂x − 1 = 0. Set u = log₂x with x > 0.
The quadratic u² − u − 1 = 0 has roots u = (1 ± √5)/2. Both are real, and neither needs to be an integer.
Return to x: the two exact answers are x = 2^[(1 + √5)/2] and x = 2^[(1 − √5)/2].
Both x-values are positive. The second exponent is negative, so that x lies between zero and one, but is still admissible. Exact exponential forms are appropriate when the problem does not request numerical approximations.
Why the substituted quadratic’s discriminant matters
For an equation u² + pu + q = 0 where u = log_bx, the discriminant D = p² − 4q determines how many distinct real u-roots exist.
Because log_b maps positive x-values bijectively onto all real u-values for valid b ≠ 1, each distinct real u-root corresponds to exactly one positive x-solution.
Thus D > 0 gives two distinct positive x-solutions, D = 0 gives one distinct positive x-solution, and D < 0 gives no real x-solutions for this simple form.
This correspondence is different from u = x² or u = b^x, where the substituted variable has a restricted non-negative or positive range and some real u-roots must be rejected.
A repeated logarithmic root gives one x-value
Take (log₂x)² − 4log₂x + 4 = 0. Substitution produces u² − 4u + 4 = (u − 2)² = 0.
The only distinct u-value is 2, even though it is a repeated root of the quadratic. Therefore the only distinct real x-solution is x = 4.
Do not list x = 4 twice or invent another logarithmic branch. For a fixed valid base b, log_b is one-to-one on its positive-input domain.
A repeated algebraic root affects multiplicity, not the number of distinct input values in a solution set. The same distinction appears in ordinary quadratic and polynomial root questions.
A negative discriminant means no real solution
Consider (log₂x)² + 2log₂x + 5 = 0. After substitution, the quadratic is u² + 2u + 5.
Complete the square: (u + 1)² + 4 = 0. Since a real square is non-negative, the left side is always at least 4 and cannot equal zero.
Therefore there is no real u and hence no real x satisfying the logarithmic equation.
There is no need to search for positive x by trial or use a calculator on a negative discriminant. The completed-square reasoning proves impossibility for the entire permitted real domain.
A parameter may control the number of solutions
Consider (log₂x)² − 4log₂x + k = 0 with x > 0 and real k. Set u = log₂x.
The quadratic u² − 4u + k = 0 has discriminant D = 16 − 4k. It has two distinct real roots when k < 4, one repeated real root when k = 4 and no real roots when k > 4.
Because every real u corresponds to one positive x, these are also the counts of distinct positive x-solutions of the original equation.
For k = 4, the repeated u-root is 2, giving x = 4. For k = 0, the roots are u = 0 and 4, giving x = 1 and 16. The sign of u itself requires no restriction.
The logarithm’s base may lie below one
Suppose (log_(1/2)x)² − 3log_(1/2)x + 2 = 0 for x > 0. Set u = log_(1/2)x.
The quadratic still gives u = 1 or 2. But returning to x uses the correct base: x = (1/2)^1 = 1/2 or x = (1/2)^2 = 1/4.
Both values are positive. The decreasing nature of the base-one-half logarithm changes the correspondence between larger u and larger x, but does not invalidate the substitution for an equality.
Inequalities require further care because converting ranges of u to ranges of x reverses order for a base below one. The base affects the final mapping even when the quadratic in u looks identical.
Extend the technique to a logarithmic inequality
Solve (log₂x)² − 3log₂x + 2 ≤ 0. Let u = log₂x and remember x > 0.
The quadratic inequality (u − 1)(u − 2) ≤ 0 holds for 1 ≤ u ≤ 2.
Since log₂ is increasing, 1 ≤ log₂x ≤ 2 is equivalent to 2¹ ≤ x ≤ 2².
Therefore the answer is 2 ≤ x ≤ 4. The endpoints are included because the original inequality allows equality and both logarithm arguments are valid. This is a direct transfer from solving an equation in u to solving a quadratic sign region in u.
A base below one reverses the interval endpoints
Now solve (log_(1/2)x)² − 3log_(1/2)x + 2 ≤ 0. The transformed quadratic again yields 1 ≤ u ≤ 2.
But u = log_(1/2)x is decreasing in x. Converting the interval gives (1/2)² ≤ x ≤ (1/2)^1.
Thus 1/4 ≤ x ≤ 1/2. The smaller x corresponds to the larger logarithm value because the base lies between zero and one.
A student who copies the base-two interval 2 ≤ x ≤ 4 or writes 1/2 ≤ x ≤ 1/4 has forgotten either the new base or the direction of the monotonic mapping.
An inequality may create two separated x-regions
Consider (log₂x)² − 3log₂x + 2 > 0. The transformed quadratic (u − 1)(u − 2) is positive when u < 1 or u > 2.
For base 2, the corresponding inputs satisfy log₂x < 1 or log₂x > 2. Since the logarithm is increasing, these become 0 < x < 2 or x > 4.
The complete solution is (0, 2) ∪ (4, ∞). The interval below 2 begins strictly above zero because the logarithm’s argument must be positive.
The two-region answer is mathematically natural: the quadratic in u is positive outside its roots, and the logarithm maps the positive x-axis continuously onto the entire real u-axis.
Use the correct logarithm power laws
The logarithm laws include log_b(MN) = log_bM + log_bN and log_b(M^p) = p log_bM, under appropriate positive-argument conditions.
They do not include log_b(M + N) = log_bM + log_bN. That false shortcut can change both the algebra and the domain of an equation.
When the original equation contains log₂(x − 1), do not split it into log₂x − log₂1. The logarithm of a difference is not generally the difference of logarithms.
Use the full argument as a single quantity under substitution. Domain checks should be based on that full argument, and valid power laws should be applied only after their input conditions are satisfied.
The substituted expression is not the original input
In the opening example, u = log₂x. Finding u = 2 does not mean x = 2; it means log₂x = 2 and therefore x = 4.
Sometimes one of the answers happens to coincide numerically, as when u = 1 yields x = 2 rather than 1. Relying on such coincidences is unreliable.
Write a clear return step after solving the quadratic: “u = …, so log_b(argument) = …, hence argument = b^(…).”
This keeps the substitution’s meaning visible and helps the student catch a skipped exponential conversion before the final answer is submitted.
An equation with different logarithm bases needs caution
The simple hidden-quadratic method assumes the same logarithmic quantity appears repeatedly. If an equation contains log₂x and log₃x, they are related but not identical.
Use change of base if appropriate: log₃x = ln x/ln 3 and log₂x = ln x/ln 2. Each is a different constant multiple of ln x.
A substitution u = ln x may then simplify the equation, provided all arguments remain positive and the constants are handled correctly.
Do not rename both different-base logarithms as one u without their scaling factors. The strength of substitution depends on accurately identifying a common underlying quantity.
A compact exam solution routine
First state the domain of every logarithm argument. Second identify a repeated logarithmic expression and substitute u for it.
Third solve the resulting quadratic, retaining all real u-values unless some additional condition restricts them. Fourth convert each u-value back through the original logarithm base and full argument.
Fifth check the x-domain and substitute candidates into the unsimplified equation. For inequalities, translate the u-intervals using the base’s monotonicity, reversing order when 0 < b < 1.
This sequence reuses familiar algebra while preserving the meaning of logarithms. It is short enough for mixed-topic examinations and clear enough to diagnose errors.
Practice questions with checked answers
Question one: (log₂x)² − 3log₂x + 2 = 0. Answer: x = 2 or 4.
Question two: (log₂x)² + log₂x − 2 = 0. Answer: x = 1/4 or 2.
Question three: (log₃(x − 1))² − log₃(x − 1) − 2 = 0. Answer: x = 4/3 or 10.
Question four: (log₂x)² − 4log₂x + 4 = 0. Answer: x = 4.
Question five: (log₂x)² − 3log₂x + 2 ≤ 0. Answer: 2 ≤ x ≤ 4.
Question six: (log_(1/2)x)² − 3log_(1/2)x + 2 ≤ 0. Answer: 1/4 ≤ x ≤ 1/2.
For every question, write the full logarithm argument’s positivity condition before making a substitution.
Parent questions and first-error diagnosis
Can a logarithm be negative? Yes. Its input must be positive, but its output can be any real number for a valid base.
Does squaring a logarithm mean taking the logarithm of a square? No. Parentheses and the placement of the exponent determine the operation.
Why is a negative u-root sometimes valid? Because u is a logarithm output, not a square or a positive exponential output.
Can all real u-values be returned to x? For u = log_bx with valid b and x > 0, yes. More complicated original arguments still need their own domain and algebraic checks.
A tutor should identify whether the child’s first difficulty is notation, quadratic factorisation, exponential conversion or domain reasoning. These lead to different targeted repairs.
Related learning and syllabus fit
Hidden-quadratic logarithm equations connect quadratic algebra, logarithm laws, inverse exponential operations and domain restrictions. Different equation and inequality variations should match the student’s taught syllabus and examination pathway.
For Singapore, consult the relevant SEAB lists for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2, as appropriate.
Related guides include How to Spot a Hidden Quadratic in an Exponential Equation, Why a Logarithmic Inequality Reverses When Its Base Is Below One, How to Track Domain Restrictions Through Composite Functions and the Additional Mathematics archive.
Families may enquire about suitable eduKate Singapore three-student Mathematics tutorials near Sixth Avenue MRT. The essential skill is recognising a familiar quadratic hidden inside logarithmic notation and returning every valid substituted result to the original variable. Properly taught kids shine a bright light into the future.