A logarithm with base between zero and one is a decreasing function. That means an inequality reverses direction when comparing its logarithm arguments or when converting it into exponential form. First require every logarithm argument to be strictly positive, then use the correct monotonicity.
For example, log base 1/2 of (x − 1) ≥ 2 means 0 < x − 1 ≤ (1/2)² = 1/4. The answer is 1 < x ≤ 5/4. Using the increasing-logarithm rule from base 2 would give the inequality in the wrong direction. This guide covers sign reasoning, change of base, two-log comparisons, boundary inclusion and checks that avoid invalid solutions.
A logarithm’s base changes whether the graph rises or falls
A real logarithmic function log_b u requires a positive base b different from 1 and a strictly positive argument u. The base divides the behaviour into two cases.
When b > 1, increasing the argument increases the logarithm. For example, log₂1 = 0, log₂2 = 1 and log₂4 = 2. This is an increasing function.
When 0 < b < 1, increasing the argument decreases the logarithm. For base 1/2, log_(1/2)(1/4) = 2, log_(1/2)1 = 0 and log_(1/2)2 = −1.
Therefore the statement “larger argument gives larger logarithm” is not universally true. It depends on the base, and that dependence controls the direction of logarithmic inequalities.
Why base one-half makes the function decrease
The logarithm y = log_(1/2) u answers the question “what exponent y makes (1/2)^y equal u?” As y increases, powers of 1/2 become smaller.
For instance, (1/2)^2 = 1/4, (1/2)^1 = 1/2 and (1/2)^0 = 1. As the argument moves from 1/4 to 1/2 to 1, the corresponding logarithm moves from 2 to 1 to 0.
This reversal is built into the exponential relationship. A decreasing exponential has a decreasing inverse function, which is its logarithm on positive inputs.
Students can remember the direction through this simple table rather than memorising an unexplained inequality rule. The operation is the same as other order-reversing functions: applying a decreasing function reverses a comparison.
The domain always comes before the inequality
For log_(1/2)(x − 1), the argument must satisfy x − 1 > 0. Therefore x > 1 before any logarithmic inequality is solved.
The condition is strict. Logarithms of zero and negative real numbers are not defined in this setting. It is invalid to include x = 1 merely because a transformed inequality uses ≤.
For log_(1/2)(2x − 1), the condition is x > 1/2. For log_(1/2)(5 − x), the condition is x < 5.
When two logarithms appear, both arguments must be positive. The final solution is the intersection of the domain conditions and the inequality derived by comparing arguments. Skipping that intersection can include false solutions even when the direction reversal was correct.
Complete example: log base one-half of x minus one
Solve log_(1/2)(x − 1) ≥ 2. Start with x > 1 from the logarithm’s domain.
Because 0 < 1/2 < 1, the logarithm decreases. A logarithm value at least 2 corresponds to an argument no larger than (1/2)^2 = 1/4.
Therefore x − 1 ≤ 1/4. Combining with the positive-argument condition gives 0 < x − 1 ≤ 1/4.
Add 1 throughout: 1 < x ≤ 5/4. The lower boundary is excluded because the logarithm argument would be zero. The upper boundary is included because at x = 5/4 the argument equals 1/4 and the logarithm equals 2.
Why x = 5/4 is the correct upper endpoint
Substitute x = 5/4 into the original expression. Then x − 1 = 1/4 and log_(1/2)(1/4) = 2, so the ≥ condition holds with equality.
Try x = 3/2. The argument is 1/2 and its logarithm is 1, which is less than 2. Thus x = 3/2 must not belong to the answer.
Try x = 9/8. The argument is 1/8, whose base-one-half logarithm is 3, which does satisfy ≥ 2. This input lies between 1 and 5/4.
These tests confirm the interval’s direction. They are checks rather than a proof of every permitted value; the decreasing-function reasoning supplies that general proof.
Contrast the same inequality with base two
Now solve log₂(x − 1) ≥ 2. The argument still requires x > 1, but base 2 is greater than one, so its logarithm is increasing.
The inequality becomes x − 1 ≥ 2² = 4. Thus x ≥ 5.
Compare this with the earlier base-one-half answer 1 < x ≤ 5/4. The two domains begin at the same boundary, yet the solution regions lie in very different places.
This pair is a useful retrieval exercise. The student should be able to explain the difference using base monotonicity, not simply recall which sign was written in one model answer.
Change of base reveals the negative divisor
Use the change-of-base formula log_(1/2)u = ln u/ln(1/2), for u > 0. Since ln(1/2) is negative, dividing by it reverses numerical order.
For log_(1/2)(x − 1) ≥ 2, write ln(x − 1)/ln(1/2) ≥ 2, with x > 1.
Multiplying by ln(1/2), a negative number, reverses the inequality: ln(x − 1) ≤ 2ln(1/2) = ln(1/4).
Because the natural logarithm is increasing, x − 1 ≤ 1/4. Combine with x − 1 > 0 to obtain 1 < x ≤ 5/4. This route proves the same result using a familiar sign rule from algebra.
A second example with a negative logarithm boundary
Solve log_(1/2)(2x − 1) < −1. The argument requires 2x − 1 > 0, so x > 1/2.
The base is decreasing, hence the logarithm being less than −1 means its argument must be greater than (1/2)^(−1) = 2.
Therefore 2x − 1 > 2, or 2x > 3. The solution is x > 3/2.
Check x = 2: the argument is 3 and log_(1/2)3 is approximately −1.585, less than −1. At x = 1, the logarithm is 0, which does not satisfy the inequality. Both checks support the strict endpoint.
A zero logarithm boundary is especially easy to check
For every valid base b, log_b 1 = 0. With a base smaller than one, inputs larger than 1 produce negative logarithms, while inputs between zero and one produce positive logarithms.
Solve log_(1/2)(x − 3) ≥ 0. The domain is x > 3. The decreasing relationship gives x − 3 ≤ 1.
Thus 3 < x ≤ 4. At x = 4 the argument is 1 and the logarithm equals zero. At x = 3 the argument is zero and the logarithm is undefined.
This example helps students internalise the base-one-half graph: positive logarithm outputs occur for small positive arguments, not for large ones.
Comparing two logarithms with the same decreasing base
Suppose log_(1/2)(x + 2) ≤ log_(1/2)(5 − x). First require x + 2 > 0 and 5 − x > 0. Thus −2 < x < 5.
Because the same logarithm function is decreasing, the comparison reverses when we remove the logarithms: x + 2 ≥ 5 − x.
Solve 2x ≥ 3, giving x ≥ 3/2. Intersect with the original domain to obtain 3/2 ≤ x < 5.
At x = 3/2 the arguments are equal, so equality is allowed. At x = 5 the second argument is zero, so the endpoint remains excluded even though the simplified algebraic comparison might seem harmless there.
Both logarithm arguments must be positive
Consider log_(1/2)(x − 2) > log_(1/2)(4x − 9). The first logarithm requires x > 2, and the second requires x > 9/4.
Together the domain is x > 9/4. The logarithm is decreasing, so the inequality reverses to x − 2 < 4x − 9.
Rearrange to 7 < 3x, giving x > 7/3. This automatically satisfies x > 9/4, so the final answer is x > 7/3.
If a student forgets the argument restrictions, a different example may leave an invalid region. Always state the domain before comparing arguments, even when one final bound happens to make the earlier conditions redundant.
Why an equality boundary may or may not be included
For log_(1/2)(x − 1) ≥ 2, the endpoint x = 5/4 is included because the logarithm is defined and equals 2 there. The endpoint x = 1 is excluded because its argument is zero.
For log_(1/2)(2x − 1) < −1, the boundary x = 3/2 gives equality −1 rather than strict inequality, so it is excluded.
Three different reasons can therefore exclude a boundary: it might violate the input domain, it might make a denominator zero during algebra, or it might fail a strict comparison.
A careful student determines which reason applies. Copying square or round brackets mechanically from the original inequality does not provide a reliable answer.
A composite logarithm can lead to a quadratic inequality
Consider log_(1/2)(x² − 5x + 6) ≥ 0. The argument must be positive: (x − 2)(x − 3) > 0, so x < 2 or x > 3.
Because the logarithm is decreasing and log_(1/2)1 = 0, the comparison gives x² − 5x + 6 ≤ 1. Thus x² − 5x + 5 ≤ 0.
The roots of x² − 5x + 5 = 0 are (5 − √5)/2 and (5 + √5)/2. The quadratic is non-positive between those roots, inclusive.
Intersect that interval with the strict positive-argument domain. The answer is [(5 − √5)/2, 2) ∪ (3, (5 + √5)/2]. The gap from 2 to 3 is excluded because the logarithm argument is non-positive there.
Why the quadratic example has two disconnected intervals
The expression x² − 5x + 6 = (x − 2)(x − 3) is positive to the left of 2 and to the right of 3. It is zero at 2 and 3 and negative between them.
The additional bound x² − 5x + 6 ≤ 1 limits how far away from those zeros the argument may rise. The permitted values therefore form a short interval just left of 2 and another just right of 3.
This is a natural result of intersecting the domain with a magnitude condition. There is no requirement that a logarithmic inequality produce one continuous interval.
A sketch of the quadratic argument can help predict this shape, but the exact endpoints come from factorisation and the quadratic formula. The graph is a check on the algebra, not a substitute for it.
Logarithmic outputs can be negative
The argument of a real logarithm must be positive, but its output may be negative. For example log_(1/2)2 = −1, while log_(1/2)(1/4) = 2.
Students sometimes mistakenly impose log_(1/2)u ≥ 0 as a general domain condition. That would remove perfectly valid arguments greater than one.
The domain condition concerns u > 0 only. An inequality such as log_(1/2)u ≥ 0 is an additional question about which valid arguments produce non-negative output, not a precondition for the function’s existence.
Separating domain restrictions from the requested output comparison is central to accurate A-Math reasoning. The same distinction appears in exponential, radical and rational functions.
Do not treat base one as valid
The standard real logarithm log_b u requires b > 0 and b ≠ 1, with u > 0. Base one is excluded because 1 raised to any real power remains 1.
There is no unique exponent that maps base one to a general positive argument. Therefore log₁u is not a valid logarithmic function in this setting.
A negative base is also outside the usual real-logarithm definition for arbitrary real exponents. The base restrictions must be checked separately from argument positivity.
When the base itself contains a parameter, determine which parameter values make it positive and different from one before interpreting the logarithmic inequality. The monotonicity case may then depend on whether the valid base lies below or above one.
An exponential inequality is the matching idea
Solve (1/2)^x < 1/8. Since 1/8 = (1/2)^3 and the exponential with base 1/2 is decreasing, the inequality becomes x > 3.
This parallels the logarithm reversal. A larger exponent produces a smaller positive output when the base is between zero and one.
For comparison, 2^x < 8 = 2³ would give x < 3 because the base-two exponential is increasing.
The same general monotonicity principle applies to a function and its inverse. Understanding it removes the need to learn unrelated sign tricks for exponential and logarithmic inequalities.
A quick number-line check
For the solution log_(1/2)(x − 1) ≥ 2, the final interval is 1 < x ≤ 5/4. Plot the two boundary values on a number line.
The point 1 is open because the logarithm is undefined there. The point 5/4 is closed because the logarithm equals 2. Shade only between them.
Test x = 9/8: x − 1 = 1/8, so the logarithm equals 3 and the inequality is true. Test x = 3/2: the logarithm equals 1 and is too small.
This short check catches the most common error, which is shading to the right of 5/4 after using the increasing-log rule instead of the decreasing-log rule.
A reliable five-stage routine
First verify the base is positive and not one. Decide whether it is greater than one or between zero and one.
Second require every logarithm argument to be strictly positive. Third apply the appropriate monotonicity: preserve the inequality for base greater than one, reverse it for base below one.
Fourth solve the resulting algebraic inequality and intersect with the logarithm domain. Fifth test any delicate endpoints in the original expression.
This procedure works for a single logarithm compared with a constant and for two same-base logarithms. Problems with different bases or additional transformations may require change-of-base reasoning, but the base sign and argument conditions remain indispensable.
Practice questions with checked answers
Question one: log_(1/2)(x − 1) ≥ 2. Answer: 1 < x ≤ 5/4.
Question two: log_(1/2)(2x − 1) < −1. Answer: x > 3/2.
Question three: log_(1/2)(x − 3) ≥ 0. Answer: 3 < x ≤ 4.
Question four: log_(1/2)(x + 2) ≤ log_(1/2)(5 − x). Answer: 3/2 ≤ x < 5.
Question five: log_(1/2)(x − 2) > log_(1/2)(4x − 9). Answer: x > 7/3.
Question six: log_(1/2)(x² − 5x + 6) ≥ 0. Answer: [(5 − √5)/2, 2) ∪ (3, (5 + √5)/2].
For every question, write argument positivity before reversing any inequality. That domain step should become automatic rather than an optional final check.
A third base: one-third instead of one-half
Solve log_(1/3)(x + 1) > −2. The domain requires x + 1 > 0, hence x > −1. Since 1/3 lies between zero and one, the logarithm is decreasing.
The inequality becomes x + 1 < (1/3)^(−2) = 9, because a larger logarithmic output corresponds to a smaller positive argument. Therefore x < 8.
Combining both conditions gives −1 < x < 8. At x = 8 the logarithm is −2, which does not satisfy the strict > inequality, while x = −1 is outside the logarithm’s domain.
This example confirms that the reversal rule is not peculiar to the base 1/2. Every real base strictly between zero and one produces a decreasing logarithm.
Adding logarithms requires the product law and a domain
Consider log_(1/2)x + log_(1/2)(x − 1) ≥ 0. Both arguments must be positive, so x > 1.
Using the logarithm product law, the left side equals log_(1/2)[x(x − 1)]. Because the base is decreasing and log_(1/2)1 = 0, the inequality requires x(x − 1) ≤ 1. The domain already makes the product positive.
Solve x² − x − 1 ≤ 0. Its roots are (1 − √5)/2 and (1 + √5)/2, and the quadratic is non-positive between them.
Intersect with x > 1. The solution is 1 < x ≤ (1 + √5)/2. The left boundary is excluded because log_(1/2)(x − 1) would be undefined at x = 1.
Do not apply the product law before checking that each original argument is valid.
A negative multiplier can cause another reversal
Solve −2log_(1/2)x ≥ 2. First write the logarithm domain x > 0.
Divide both sides by −2. Since this number is negative, reverse the inequality to obtain log_(1/2)x ≤ −1. Now use the fact that the logarithm itself is decreasing, which reverses the comparison with its argument: x ≥ (1/2)^(−1) = 2.
Thus the final solution is x ≥ 2. Check x = 2: log_(1/2)2 = −1, and multiplying by −2 gives 2, so equality is permitted.
There were two separate inequality reversals: one from dividing by the negative coefficient and another from applying the decreasing logarithm. Students should show them as two distinct steps rather than cancelling signs informally.
Different logarithm bases cannot simply be stripped
Suppose the problem is log_(1/2)x ≥ log₂x. The arguments are both x, and the domain is x > 0, but the bases are different.
Use change of base: log_(1/2)x = −log₂x, because ln(1/2) = −ln 2. Let L = log₂x. The inequality becomes −L ≥ L, so −2L ≥ 0 and L ≤ 0.
Since base 2 is increasing, log₂x ≤ 0 gives x ≤ 1. Combine with x > 0 to obtain 0 < x ≤ 1.
A tempting shortcut of deleting both logarithms would compare x with itself and claim every positive x works. That fails because the logarithms are different functions. Convert them to one common base before comparing their values.
Graph behaviour provides a reliable sign check
The graph y = log_(1/2)x is defined only for x > 0. It passes through (1, 0), rises without bound as x approaches zero from the right, and falls without bound as x increases without limit.
The points (1/4, 2), (1/2, 1), (1, 0) and (2, −1) provide an easy exact sketch. Reading these points in increasing x-order, the y-values steadily decrease.
Therefore log_(1/2)x ≥ 2 must correspond to very small positive x-values, at most 1/4. Similarly log_(1/2)x < −1 requires x > 2.
This qualitative check is useful after algebra. If a final inequality claims that a larger positive argument produces a larger base-one-half logarithm, revisit the direction reversal before submitting the answer.
When a parameter controls the logarithm base
Some enrichment questions use a parameter in the base, such as log_(k−1)x. Before deciding whether the logarithm increases or decreases, determine which values of k make a valid base.
The base must satisfy k − 1 > 0 and k − 1 ≠ 1. Therefore k > 1 and k ≠ 2.
For 1 < k < 2, the base lies strictly between zero and one, so the logarithm is decreasing. For k > 2, the base is greater than one, so the logarithm is increasing.
The value k = 2 gives base one and is invalid. Values k ≤ 1 make the base non-positive and invalid in the standard real logarithm. A parameter-dependent inequality must be solved by cases only after this base-domain classification has been established.
Typical mistakes and how to repair them
The most frequent mistake is treating log_(1/2) as increasing. A three-value table using arguments 1/4, 1 and 2 can establish the correct direction more clearly than repeating a slogan about reversing signs.
Another student knows the direction but forgets that arguments must be positive. A changed problem with two logarithms and different argument restrictions is a useful repair. A third may have the correct inequality but include an argument-zero endpoint.
A fourth may apply a negative-base logarithm or use a calculator without checking whether the selected base is defined. These are domain and notation problems that need targeted practice.
A tutor should identify the first broken decision, then give an independent variation where the same principle appears with different constants or a different log argument.
Syllabus guidance and connected reading
Logarithms, inequalities and inverse functions connect through the monotonicity of exponential functions. The exact types of log inequalities covered or examined depend on the student’s syllabus year and school sequence. Bases below one can also be valuable enrichment when a particular paper focuses on other forms.
Singapore students should refer to the appropriate SEAB documents for 2026 GCE O-Level Additional Mathematics 4049, 2027 SEC G3 K341 or 2027 SEC G2 K232.
Related guides include How to Track Domain Restrictions Through Composite Functions, How to Spot a Hidden Quadratic in an Exponential Equation and How to Disprove a False Algebraic Rule with One Counterexample. See the Additional Mathematics archive or enquire about eduKate Singapore small-group tutorials near Sixth Avenue MRT.
The aim is to know exactly why the inequality direction changes and why only positive logarithm arguments can appear in the answer. Properly taught kids shine a bright light into the future.