Question 104689
Some rules to remember 
{{{k log (x)=log((x^k))}}}
{{{log((x))+log((y))=log((xy))}}}  
{{{log((x))-log((y))=log((x/y))}}}  
Now let's look at your example.
{{{log((7y + 1)) = 2log((y+3))-log((2))}}}
{{{log((7y + 1)) - 2log((y+3))= -log((2))}}} Bring the variable expressions to one side. 
{{{log((7y + 1)) - log((y+3)^2)= log((2^(-1)))}}} Power rule from above both sides.
{{{log(((7y + 1)/(y+3)^2))= log((2^(-1)))}}} Division rule from above.
{{{(7y + 1)/(y+3)^2= 1/2}}} Inverse function both sides. 
{{{2(7y + 1)=(y+3)^2}}} Multiply both sides by {{{2*(y+3)^2}}}
{{{14y+2=y^2+6y+9}}} Distribute and expand. 
{{{y^2-8y+7=0}}}Group all terms on one side.
{{{(y-1)(y-7)=0}}} Factor the equation.
{{{y=1}}} and {{{y=7}}}. Two solutions. 
Verify your answer
{{{log((7y + 1)) = 2log((y+3))-log((2))}}}
{{{log((7(1) + 1)) = 2log((1+3))-log((2))}}}
{{{log((8))=2log((4))-log((2))}}}
{{{log((8))=log((4^2/2))}}} Power and division rule, right hand side. 
{{{log((8))=log((8))}}}
Good answer. 
{{{log((7y + 1)) = 2log((y+3))-log((2))}}}
{{{log((7(7) + 1)) = 2log((7+3))-log((2))}}}
{{{log((50))=2log((10))-log((2))}}}
{{{log((50))=log((10^2/2))}}} Power and division rule, right hand side. 
{{{log((50))=log((50))}}}
Good answer.