Logarithmic Inequalities: Follow the Base, Keep the Domain

A smaller base can turn the order around

Start with two positive numbers, $u<v$. If the base $b>1$, the logarithm increases: the larger input has the larger output. If $0<b<1$, it decreases: the larger input has the smaller output.

These are statements about a function’s order, not rules about moving a written base to another side. Think of two points on the curve. For an increasing curve, reading from left to right moves upward. For a decreasing curve, it moves downward. In both cases, the inputs must stay positive.

Comparing a logarithm with a number uses the same reasoning. Write $k=\log_b(b^k)$. Then $\log_b U<k$ compares two logarithms with the same base. For $b>1$, it becomes $U<b^k$, while for $0<b<1$, it becomes $U>b^k$. Retain $U>0$ in either case.

The words “larger than 1” describe the base, not the input. For instance, base 2 preserves order even when both inputs lie between 0 and 1. A negative logarithm output is also perfectly valid.

Use the activity to predict which output is higher before revealing the values. Then switch between bases 2 and 1/2 while keeping the two inputs unchanged. The order reversal should make sense visually and in the exact numbers.

An algebraic interval is only part of the answer

For a fixed base above 1, solving the comparison is usually straightforward. The trap is forgetting that every original logarithm still needs a positive argument. An answer such as $x<2$ might include many values for which the question is not even defined.

Use two separate lines of reasoning. First find the domain, by making each argument positive. Then solve the inequality using monotonicity. Finally take their intersection: the values that belong to both sets. A small number line can keep this last step visible.

For $\ln(2x-1)<\ln(x+1)$, the first argument requires $x>\tfrac12$, and the second requires $x>-1$. Together they reduce to $x>\tfrac12$. Since $\ln$ has base $e>1$, comparison of the arguments gives $x<2$. Both statements must hold, producing a bounded interval.

Endpoint symbols carry information. A strict comparison usually excludes an equality endpoint. A non-strict comparison may include it, but only if the original expressions exist there. An argument equal to zero is never allowed, even if the final inequality sign is $\le$ or $\ge$.

For example, $\log_2(3x-1)\le4$ gives $3x-1\le16$, while the domain gives $3x-1>0$. Together they say $\tfrac13<x\le\tfrac{17}3$. The left endpoint is open and the right endpoint is closed for different reasons. Test an endpoint in the original expression whenever you are uncertain.

Reverse the comparison, not the domain

With a base between 0 and 1, the logarithm decreases. This changes the comparison between the arguments, but it does not change the requirement that each original argument be positive. Treat these as two separate facts.

For $\log_{1/9}(x+1)\ge0$, rewrite the right side as $\log_{1/9}1$. A greater or equal logarithm output must come from a smaller or equal input, so $x+1\le1$. The original argument still requires $x+1>0$. Together these inequalities describe $0<x+1\le1$.

A common mistake is to stop at $x\le0$. That includes numbers such as−2, where the argument $x+1$ is negative. Another mistake is to reverse the positivity condition because the base is small. The domain condition is simply part of the definition of a real logarithm; it is not a comparison being passed through the logarithmic function.

Negative thresholds are often friendly. Since $(1/2)^{-1}=2$, the inequality $\log_{1/2}U<-1$ means $U>2$. Think in powers first, then use the decreasing direction. You do not need a decimal approximation.

After finding the interval, try one value inside it and one tempting value outside it. The first should make the original comparison true. The second should either make the comparison false or expose an undefined logarithm. Numerical tests help detect a slip; the domain and monotonicity reasoning establishes the whole solution set.

When the base is the unknown, make cases

A fixed base tells you immediately whether to preserve or reverse order. In $\log_x(x+1)>1$, the base itself is the unknown. You must first decide which family the base belongs to: $0<x<1$, or $x>1$. The values $x\le0$ and $x=1$ are excluded from the start.

Write the original domain before splitting. Here $x>0$, $x\ne1$ already make $x+1>0$. In the increasing-base case $x>1$, compare $x+1$ with $x^1$ without reversing. In the decreasing-base case $0<x<1$, reverse the comparison.

A reduced statement such as $1>0$ is true, but it does not mean the original problem accepts every real number. It means every number within the current case and original domain passes that comparison.

Use this reusable structure:

• Within each case, intersect the domain, the case restriction and the algebraic result.
• After completing the cases, unite their valid solution sets.

Intersection means “all these conditions at once”; union means “this case or that case”. A case with no values contributes the empty set, $\varnothing$.

Sometimes the argument condition removes a case before any algebra. For $\log_x(x-5)$, positivity requires $x>5$, so the base is automatically above 1. Writing an unnecessary decreasing-base case can obscure that useful shortcut. Let the domain simplify your work whenever possible.

A quadratic argument can leave two separate intervals

When the argument is quadratic, the domain itself may have two pieces. For $\log_2(x^2-1)$, positivity requires $x^2-1>0$, or $|x|>1$. Both sufficiently positive and sufficiently negative values of $x$ are allowed.

Now compare the logarithm with zero. Since the base 2 is increasing, $\log_2(x^2-1)<0$ means $x^2-1<1$. Combining this with positivity gives the compact double inequality

Thus $1<x^2<2$, or $1<|x|<\sqrt2$. The absolute value describes distance from zero, so the answer has a positive interval and a matching negative one.

On a number line, place the boundaries in their correct order: $-\sqrt2<-1<1<\sqrt2$. Remember that $\sqrt2$ is about 1.414, not 2. Keep exact radicals in the final answer so the endpoint meaning is not changed by rounding.

Check the two kinds of boundary separately. At $x=\pm1$, the argument is zero and the logarithm does not exist. At $x=\pm\sqrt2$, the argument is 1 and the logarithm is 0; the strict comparison excludes it. If the question used a non-strict sign, only this second kind of endpoint could become included.

The same method works for other quadratic arguments: find where the original argument is positive, solve the comparison, and intersect. Do not automatically require $x>0$ when the logarithm contains $x^2-1$.