Question 8613
I'm assuming your problem is 
{{{log((3x)) + log((x+5)) = log((x^2+4x))}}} 


Using the second law of logarithms:
{{{log(((3x)*(x+5))) = log((x^2+4x))}}}


Since the log(M) = log(N) implies that M=N,
{{{3x(x+5) = (x^2 + 4x)}}}
{{{3x^2 + 15x = x^2 + 4x}}}, which is a quadratic equation.  


Set equal to zero,
{{{2x^2 + 11x = 0}}} and solve by factoring:


x(2x+11) = 0
x=0 or x = {{{-11/2}}}


However, both answers must be rejected, since you cannot have a log of a negative or zero number.  There is NO SOLUTION!! 


Does anyone see any errors here?


R^2 at SCC