Log3 (x+2) /=(больше или равно) 2 log1/3 (x-1)>/=-2 log2/3 (2-5x) < - 2

To solve this system of inequalities, we must first establish the domain (ODZ) where all logarithmic functions are defined, and then solve the inequalities step-by-step. 1. Determining the Domain (ODZ) For a logarithm ${\log }_b\left(a\right)$ to exist, the argument $a$ must be strictly positive ( $a>0$).

1. $x+2>0⟹x>-2$ $x-1>0⟹x>1$ $2-5x>0⟹5x<2⟹x<0.4$

Intersection of conditions: The conditions $x>1$ and $x<0.4$ are contradictory. There is no value of $x$ that is simultaneously greater than 1 and less than 0.4. 2. Analysis of the System The expression provided consists of several parts. Let's break them down to see if any part has a solution. Part A: ${\log }_3(x+2)\ge 2{\log }_{1/3}(x-1)$ Using the base change formula ${\log }_{1/3}\left(a\right)=-{\log }_3\left(a\right)$: ${\log }_3(x+2)\ge -2{\log }_3(x-1)$ ${\log }_3(x+2)+{\log }_3(x-1{)}^2\ge 0$ ${\log }_3\left(\right(x+2\left)\right(x-1{)}^2)\ge 0$This requires $x>1$ from the argument $(x-1)$. Part B: $-2{\log }_{2/3}(2-5x)<-2$ Divide by $-2$ (and flip the inequality sign): ${\log }_{2/3}(2-5x)>1$Since the base $2/3$ is less than 1, we flip the inequality sign again when removing the log: $2-5x<(2/3{)}^1$ $2-2/3<5x$ $4/3<5x⟹x>4/15\approx 0.266$This requires $x<0.4$ from the argument $(2-5x)$. 3. Conclusion To find the final solution, we must find the intersection of the requirements from all parts of the expression:

• From the first part: $x>1$ From the second part: $x<0.4$

Because the interval $(1,+\infty )$ and the interval $(-\infty ,0.4)$ do not overlap, there is no value of $x$ that satisfies the entire chain of inequalities. Final Answer: No solution (The set of solutions is empty: $\varnothing$). Would you like me to double-check the signs or constants in the original equation to see if a small typo changed the outcome?