Logarithmic inequalities with a fractional base

Your homework asks you to solve log base 1/2 of (x − 1) ≥ −2. You turn −2 into the logarithm of 4, write x − 1 ≥ 4, and hand in x ≥ 5. When you substitute x = 5, you get equality. That feels like confirmation until you try x = 4, which also satisfies the original problem. Then x = 6, which your interval includes, fails. How can the boundary be right if the interval points the wrong way?

Start logarithmic inequalities by writing the domain. The mistake happens at the comparison between two logarithms of the same base. With a base of 1/2, increasing the argument decreases the output. To work through the problem, keep two separate lines on your page. One records where the original expression exists. The other records where the comparison holds. You will need both lines even when the arithmetic seems simple enough to do mentally.

Compare outputs by putting them on one scale

Write −2 as log base 1/2 of 4. The two sides now use the same function. To see its direction, look at three inputs you can evaluate exactly: 1, 2 and 4 have outputs 0, −1 and −2, respectively. Each doubling of the argument lowers the output by one. The function is strictly decreasing over its whole positive domain, so the same ordering applies even when the two inputs are close together or are not integer powers of two.

We want the output at x − 1 to be at least the output at 4. That requires x − 1 ≤ 4. Add one to both sides to get x ≤ 5. Adding one shifts both quantities equally, so the direction stays as it is. Combine x ≤ 5 with x > 1 to get (1, 5]. The open endpoint comes from the domain. The closed endpoint comes from the equality permitted by the original comparison.

If you want a quick reality check, try x = 4 and x = 6. At x = 4 the argument is 3, which is less than 4, so the logarithm exceeds −2. At x = 6 the argument is 5, which is greater than 4, so the logarithm falls below −2. Both substitutions are in the domain. Testing points to either side gives information that testing x = 5 alone cannot provide. The proof that the whole interval works still comes from monotonicity and the domain.

Change the base and watch the interval move

Consider log₂(x − 1) ≥ 2. The domain remains x > 1, and the comparison still uses the argument 4, since 2² = 4. This time the same sample arguments 1, 2 and 4 give 0, 1 and 2. The logarithm increases. Therefore x − 1 ≥ 4 and the solution is [5, ∞). The location of the equality boundary stayed fixed while the direction changed.

There are two separate details to watch when switching bases. First, the exponent representing the same argument may change: at argument 4 it was −2 for base 1/2 and 2 for base 2. Second, the increasing or decreasing behavior determines the comparison. When the base is above one, the order of positive arguments agrees with the order of outputs. When the base is between zero and one, the orders are opposite. The cases b ≤ 0 and b = 1 are outside this ordinary real logarithm method.

This comparison also explains why the words “flip the sign” can be ambiguous. We reverse the inequality symbol when passing from outputs to inputs of a decreasing function. The signs of the inputs themselves stay as written. In particular the inputs to the logarithms must still be positive. Keep the base, the sign of the exponent and the orientation of the comparison separate when you explain your answer.

Ask Eqora about the line that changed

If your homework explanation jumps from logarithms straight to an interval, capture the original line in Eqora with the fractional base, the parentheses and the bar of ≥ visible. Compare the recognized problem with your worksheet before using any steps. Losing the denominator changes the base. Cropping off the bar turns a closed boundary into an open one. Moving the parenthesis changes the expression whose positivity you must check.

Read one step at a time and annotate what justifies it. Beside −2 becoming the logarithm of 4, write the power calculation. Beside the reversed comparison, write “decreasing on positive arguments.” Beside the final interval, write the two conditions it combines. If a step leaves you unsure, ask “Why does the comparison become x − 1 ≤ 4, and where is x > 1 used?” That gives you two specific claims to check in the explanation.

You can then ask what would change if the original symbol were strict. The inequality step would give x − 1 < 4. Both endpoints would be open for different reasons. Check that answer against the original task before copying any notation. Eqora can help you inspect a step and formulate a follow-up. The reasoning, verification and written solution are still your work, and an app response gives no guarantee of correctness or grades. The first-principles derivative exercise gives another example of keeping a restriction visible through an algebraic step.

Keep both arguments in view

Next consider log base 1/3 of (2x − 1) > log base 1/3 of (x + 2). Now both arguments depend on x. Start with 2x − 1 > 0 and x + 2 > 0. The resulting conditions are x > 1/2 and x > −2. To satisfy both you need x > 1/2. Even though the second condition adds no restriction here, checking it shows why the whole domain is valid.

The base 1/3 is between zero and one. Comparing outputs with > therefore gives 2x − 1 < x + 2. Subtract x and add one to obtain x < 3. Intersect with the domain to get (1/2, 3). At x = 3, both arguments are 5 and the comparison is strict, so equality is insufficient. At x = 1/2, the first argument is zero, so evaluating the original problem is impossible. You exclude both endpoints for distinct reasons.

Try x = 1, which produces arguments 1 and 3 and outputs 0 and −1. This point belongs to the solution. At x = 4 the arguments are 7 and 6, so the output on the left is smaller than the one on the right. This point lies in the domain but outside the solution. Keeping the two descriptions separate is particularly helpful on a number line: first shade what exists, then identify the part of that shading that satisfies the comparison.