Why Logarithmic Inequalities Need Two Checks
A logarithmic inequality is not solved by algebra alone. Two facts must be handled together: the logarithm’s base determines whether the inequality direction stays the same or reverses, and every logarithm requires a positive argument. Ignoring either fact can add values where the expression is undefined or remove values that satisfy the inequality.
For example, the expression \(\log_2(x-3)\) is defined only when \(x-3>0\), so its domain is \(x>3\). A result such as \(x<7\) is not yet a complete solution: it must be intersected with this domain.
Logarithm Domains and Monotonicity
For a real base \(b\), the logarithm \(\log_b(u)\) is defined when \(b>0\), \(b\ne 1\), and \(u>0\). Its value is the exponent \(t\) such that \(b^t=u\). Thus, if \(u\) depends on \(x\), the condition \(u(x)>0\) is part of the problem, not an optional check.
The direction in which \(\log_b(u)\) changes with \(u\) depends on \(b\). When \(b>1\), the exponential function \(b^t\) is increasing, and its inverse \(\log_b(u)\) is increasing. When \(0<b<1\), \(b^t\) is decreasing, so its inverse is decreasing.
The rule for comparing a logarithm with a constant follows from the same fact. If \(c\) is real, then \(\log_b(u)<c\) compares \(u\) with \(b^c\): for \(b>1\), it gives \(u<b^c\); for \(0<b<1\), it gives \(u>b^c\). In either case, \(u>0\) must also hold.
Why the Inequality Direction Changes
A logarithm is the inverse of its exponential function. If \(b>1\), the exponential function preserves order: \(s<t\) implies \(b^s<b^t\). Its inverse therefore preserves order as well. In particular, if \(\log_b(u)<\log_b(v)\), applying the increasing exponential function gives \(u<v\).
If \(0<b<1\), the exponential function reverses order: \(s<t\) implies \(b^s>b^t\). Applying it to \(\log_b(u)<\log_b(v)\) therefore gives \(u>v\). That is the reason the inequality reverses for these bases. In both cases, this reasoning is valid only for positive \(u\) and \(v\), where the logarithms are defined.
A Reliable Solving Process
Use the following sequence whenever a variable appears inside a logarithm. If there is more than one logarithm, write down every argument restriction before simplifying or comparing expressions. After solving the resulting inequality, intersect its solution with all those restrictions.
For each logarithm, require its entire argument to be strictly positive.
Check whether it is greater than one or between zero and one.
Convert the logarithmic inequality using the monotonicity rule, reversing the direction for a base between zero and one.
Keep only values satisfying every domain condition, and check boundary behavior and representative values.
Worked Inequalities
Worked Example: A Logarithm with Base Greater Than One
Solve \(\log_2(3x-1)\leq 2\).
Step 1: Find the domain. The argument must be positive:
Step 2: Use the base. Since \(2>1\), the logarithm is increasing. Also, \(2= \log_2(4)\), because \(2^2=4\). Therefore, within the domain,
Step 3: Solve and intersect.
Combining \(x\leq \frac{5}{3}\) with \(x>\frac{1}{3}\) gives \(\left(\frac{1}{3},\frac{5}{3}\right]\).
Check: At \(x=1\), the argument is \(3(1)-1=2>0\), and \(\log_2(2)=1\leq2\). At \(x=\frac{5}{3}\), the argument is \(3(\frac{5}{3})-1=4\), so \(\log_2(4)=2\), and the endpoint is included. At \(x=\frac{1}{3}\), the argument is zero, so that endpoint is excluded because the logarithm is undefined.
Worked Example: A Logarithm with Base Between Zero and One
Solve \(\log_{\frac{1}{3}}(x+2)>-1\).
Step 1: Find the domain.
Step 2: Use the base and convert the right side. The base \(\frac{1}{3}\) lies between zero and one, so the logarithm is decreasing. Also, \(-1=\log_{\frac{1}{3}}(3)\), since \((\frac{1}{3})^{-1}=3\). Thus the inequality reverses when comparing arguments:
Step 3: Solve and intersect.
Combining this with \(x>-2\) gives \((-2,1)\).
Check: At \(x=0\), the argument is \(2\), and \(\log_{\frac{1}{3}}(2)>-1\) because \(2<3\) and the logarithm is decreasing. At \(x=1\), the argument is \(3\), giving \(\log_{\frac{1}{3}}(3)=-1\); equality does not satisfy the strict inequality, so \(1\) is excluded. At \(x=-2\), the argument is zero, so the logarithm is undefined.
Worked Example: Two Logarithms and Their Shared Domain
Solve \(\log_3(x-1)+\log_3(x+1)\geq 2\).
Step 1: Require both arguments to be positive.
The first condition gives \(x>1\), and the second gives \(x>-1\). Both must hold, so the domain is \(x>1\).
Step 2: Combine the logarithms. Both logarithms have the same base and both arguments are positive on the domain. The product rule gives
Since \(3>1\), the logarithm is increasing. Since \(2=\log_3(9)\), the inequality becomes
Step 3: Solve and apply the domain.
The last inequality holds when \(x\leq-\sqrt{10}\) or \(x\geq\sqrt{10}\). Intersecting those intervals with the domain \(x>1\) removes the negative interval and leaves \([\sqrt{10},\infty)\).
Check: At \(x=\sqrt{10}\), both arguments are positive and their product is \((\sqrt{10}-1)(\sqrt{10}+1)=10-1=9\). Thus the left side is \(\log_3(9)=2\), so the endpoint is included. At \(x=2\), the arguments are \(1\) and \(3\), and the sum is \(0+1=1<2\). At \(x=4\), the arguments are \(3\) and \(5\); their product is \(15>9\), so the increasing logarithm gives a value greater than \(2\). These checks agree with the solution set.
Domain Restrictions Must Survive Simplification
Logarithm rules can simplify an inequality, but they do not erase the restrictions that made the original expression meaningful. In the two-logarithm example, combining the terms produces a single logarithm whose argument is \((x-1)(x+1)\). That product can be positive for some \(x<-1\), but those values do not belong to the original domain: each original argument must be positive separately.
For that reason, record the original domain before applying logarithm laws. When combining \(\log_b(u)+\log_b(v)\), the product rule applies when \(u>0\) and \(v>0\). It is not enough that the product \(uv\) be positive; two negative factors also have a positive product, but neither original logarithm would be defined.
A Common Mistake: Forgetting to Reverse the Inequality
Consider \(\log_{\frac{1}{2}}(x+1)\geq -1\). A common incorrect step is to treat the logarithm as increasing and write \(x+1\geq 2\), leading to \(x\geq1\). This loses valid solutions because the base is less than one.
The correct domain condition is \(x+1>0\), or \(x>-1\). The base \(\frac{1}{2}\) is between zero and one, so the logarithm is decreasing. Since \(-1=\log_{\frac{1}{2}}(2)\), the correct comparison reverses:
Intersecting with \(x>-1\) gives \((-1,1]\). For instance, \(x=0\) is in this interval: the argument is \(1\), and \(\log_{\frac{1}{2}}(1)=0\geq-1\). The incorrect answer \(x\geq1\) would wrongly discard this valid value. The other endpoint, \(x=-1\), remains excluded because it makes the logarithm’s argument zero.
Final Checks Before Writing the Answer
A solution set is complete only when it respects the original logarithms as well as the inequality manipulation. Before finalizing an answer, check these points:
- Every logarithm has a valid base: \(b>0\) and \(b\ne1\).
- Every original argument is strictly positive.
- The inequality direction matches whether \(b>1\) or \(0<b<1\).
- Any endpoint satisfies the original inequality and lies in the domain.
- The final set is the intersection of the comparison’s solutions with all domain restrictions.
The endpoint checks matter particularly for strict inequalities and logarithm domains. An endpoint produced by an equality may be included for a non-strict inequality, but an endpoint where a logarithm argument is zero is never included.
Check Your Understanding
For each question, remember to state the domain before giving the final solution set.
- Solve \(\log_5(2x+1)<2\). Which endpoint is excluded by the logarithm’s domain?
- Solve \(\log_{\frac{1}{4}}(x-3)\leq -1\). Why does the comparison direction change?
- For \(\log_2(x)+\log_2(x-2)>3\), what two domain conditions must be imposed before combining the logarithms?
- Why is it not sufficient to require \((x-1)(x+1)>0\) when the original expression is \(\log_3(x-1)+\log_3(x+1)\)?
- When comparing \(\log_b(u)\) with \(\log_b(v)\), what must be true about \(u\) and \(v\) before using monotonicity?
