Use the log addition rule:
#log_color(green)a(color(red)x)+log_color(green)a(color(blue)y)=log_color(green)a(color(red)xcolor(blue)y)#
Now combine the two #log#s, then rewrite #1# as #log(10)#, then cancel the logs on either side:
#log(x)+log(x+3)=1#
#log(x*(x+3))=1#
#log(x^2+3x)=1#
#log(x^2+3x)=log(10)#
#color(red)cancel(color(black)log)(x^2+3x)=color(red)cancel(color(black)log)(10)#
#x^2+3x=10#
#x^2+3x-10=0#
#(x+5)(x-2)=0#
#x=2,-5#
Plug in the answers to the original equation and see if they still work:
#log(x)+log(x+3)=1#
Testing #2#:
#log(2)+log(2+3)=1#
#log(2)+log(5)=1#
#0.30102...+0.69898...=1#
#1=1qquadqquad color(lightgreen)sqrt#
The solution #2# works. Testing #-5#:
#log(-5)+log(-5+3)=1#
Since #log(-5)# is undefined, this solution doesn't work. Therefore, the only solution is #x=2#.
