0 \\ x<5 \\ log_2(5-x)=log_21 \\ 5-x=1 \\ 4 = x \\ \\ 2. \ log_3(2x+1)=log_3(4-2x) \\ \left \{ {{2x+1>0} \atop {4-2x>0}} \right. \Rightarrow \left \{ {{x>-\frac{1}{2} } \atop {x<2}} \right. \Rightarrow x \in (-\frac{1}{2};2) \\ log_3(2x+1)=log_3(4-2x) \\ 2x+1=4-2x \\ 4x=3 \\ x=0,75" alt="1. \ log_2(5-x)=0 \\ 5-x>0 \\ x<5 \\ log_2(5-x)=log_21 \\ 5-x=1 \\ 4 = x \\ \\ 2. \ log_3(2x+1)=log_3(4-2x) \\ \left \{ {{2x+1>0} \atop {4-2x>0}} \right. \Rightarrow \left \{ {{x>-\frac{1}{2} } \atop {x<2}} \right. \Rightarrow x \in (-\frac{1}{2};2) \\ log_3(2x+1)=log_3(4-2x) \\ 2x+1=4-2x \\ 4x=3 \\ x=0,75" align="absmiddle" class="latex-formula">