We assume you want to find values of x that make the expression true.
Use the rules of logarithms to rewrite as a single log. Then take the antilog and solve the resulting quadratic.
[tex]\log_{6}(2x+1)=\log_{6}(x-3)+\log_{6}(x+5)=\log_{6}((x-3)(x+5))\\\\2x+1=(x-3)(x+5)=x^{2}+2x-15\\\\0=x^{2}-16=(x-4)(x+4)[/tex]
The term [tex]\log_{6}(x-3)[/tex] is only defined for x > 3, so the only solution to this equation is x = 4.
