## Precalculus (6th Edition) Blitzer

We solve as follows: $$\log_2(2x+1) -\log_2(x-2)=1 \quad \Rightarrow \quad \log_2 \left ( \frac{2x+1}{x-2} \right ) = \log_22 \quad \Rightarrow \quad \frac{2x+1}{x-2}=2 \quad \Rightarrow \quad 2x+1=2(x-2), \quad (x \neq 2) \quad \Rightarrow \quad 1=-4$$Since we reached a wrong result, we conclude that there exists no $x$ satisfying the original equation.