After having taught logarithmic equations and inequalities, the teacher assigned students a more challenging problem to solve. Students were given the task to solve the following logarithmic inequality: $$ \log x \cdot \log_5(3x - 1) \geq 0 $$ The whole step-by-step solution was written down on the board based on the students’ suggestions.
(1) All students realized that logarithms can only be applied to positive real numbers. Therefore, they added the domain restriction to the inequality: $$ \begin{aligned} x > 0 &\land 3x - 1 > 0 \cr x > 0 &\land x > \frac13 \cr x &\in \left(\frac13 ; \infty \right) \end{aligned} $$
(2) One student suggested solving the inequality as a product-type inequality, meaning that the inequality: $$ \log x \cdot \log_5(3x - 1) \geq 0 $$ is true if and only if both factors are nonnegative, i.e., $\log x \geq 0$ and $\log_5(3x - 1) \geq 0$, or both factors are nonpositive, i.e., $\log x \leq 0$ and $\log_5(3x - 1) \leq 0$.
(3) After that, students suggested splitting the problem into two separate cases. The following step was solving the first set of logarithmic inequalities: $$ \begin{aligned} \log x \geq 0 &\land \log_5(3x - 1) \geq 0 \cr x \geq 10^0 &\land 3x - 1 \geq 5^0 \cr x \geq 1 &\land 3x - 1 \geq 1 \cr x \geq 1 &\land x \geq \frac23 \end{aligned} $$ Considering the domain restriction, they obtained: $$x \in [ 1; \infty )$$
(4) In the next step, they solved the second set of logarithmic inequalities: $$ \begin{aligned} \log x \leq 0 &\land \log_5(3x - 1) \leq 0 \cr x \leq 10^0 &\land 3x - 1 \leq 5^0 \cr x \leq 1 &\land 3x - 1 \leq 1 \cr x \leq 1 &\land x \leq \frac23 \end{aligned} $$ Considering the domain restriction, they obtained: $$x \in \left( \frac13 ; \frac23\right]$$
(5) The final solution is the intersection of the intervals found in steps (3) and (4), namely intervals $[ 1; \infty )$ and $(\frac13 ; \frac23]$. Since no real number lies within the intersection of these intervals, the given logarithmic inequality has no solution.
Is their solution correct? Explain.
No, there is a mistake in step (1). The domain restriction should be: $$ \begin{gather} x(3x - 1) > 0 \cr x \in (-\infty , 0) \cup \left(\frac13 ; \infty \right) \end{gather} $$
No, there is a mistake in step (2). This procedure is incorrect.
No, there is a mistake in step (3). It should be $x \in [ \frac23 ; \infty )$.
No, there is a mistake in step (4). It should be $x \in (\frac13 ; 1]$.
No, there is a mistake in step (5). The final solution should be the union of the intervals obtained in steps (3) and (4), not their intersection.
Yes, all steps are correct.
Steps (1)-(4) are correct. A mistake occurred in step (5), where students should have made the union of the intervals from steps (3) and (4). This would result in: $$ x \in \left(\frac13 ; \frac23 \right] \cup \left[ 1; \infty \right)\cdot $$