Logarithmic inequality with base less than 1

Statement

Solve $\log_{1/3}(2x+1) \ge -2$.

Solution

EC: $2x + 1 > 0 \iff x > -\tfrac{1}{2}$.

Uniform rewriting: $-2 = \log_{1/3}(3^2) = \log_{1/3} 9$. Hence $\log_{1/3}(2x+1) \ge \log_{1/3} 9$.

Base $<1$, inequality reversed: $2x + 1 \le 9 \iff x \le 4$.

Intersection with the EC: $\boxed{-\tfrac{1}{2} < x \le 4}$.

Links

Topics: Logarithmic function
Concepts: Existence conditions · Logarithmic inequalities · Logarithm
Functions: Logarithmic function
Skills: Solving inequalities
Exercise type: Solving an inequality