ОДЗ :

[tex]\displaystyle\bf\\\left \{ {{x+1 > 0} \atop {x > 0}} \right. \ \ \ \Rightarrow \ \ \left \{ {{x > -1} \atop {x > 0}} \right. \Rightarrow \ \ x > 0\\\\\\\log_{2} (x+1)=\log_{2}( 3x)\\\\x+1=3x\\\\x-3x=-1\\\\-2x=-1\\\\x=0,5\\\\Otvet \ : \ x=0,5[/tex]