Question 15879
Ir may help to rewrite the equation:

2log(x+3) = log(x+1) + log(2x+3) 
{{{log((x+3))^2 = log ((x+1)(2x+3))}}}


Since log A = log B means that A=B, in the same way
{{{log((x+3))^2 = log ((x+1)(2x+3))}}} means that {{{(x+3)^2 = (x+1)*(2x+3) }}}
{{{x^2 + 6x +9 = 2x^2 +5x + 3}}}
{{{0 = x^2-x-6}}}
{{{0 = (x-3)(x+2) }}}
x= 3 or x= -2


You must reject x=-2, since it gives you the log of a negative, which is not defined in the resl numbers.  The final answer is x=3.


R^2 at SCC