Use the definition of a logarithm:
#color(white)=>log_color(red)a(color(green)b)=color(blue)xqquad<=>qquadcolor(red)a^color(blue)x=color(green)b#
In this case, #a# is #6#, #b# is #x+4#, and #x# is #log_6(7)#. It sounds kind of confusing, but I'll use colors so it's easier to see:
#color(white)=>log_color(red)6(color(green)(x+4))=color(blue)(log_6(7))#
#=>color(red)6^color(blue)(log_6(7))=color(green)(x+4)#
Now, the #6# and #log_6# cancel out:
#color(white)=>color(red)cancel(color(black)6)^(color(red)cancel(color(black)(log_6))(7))=x+4#
#=>7=x+4#
#color(white)=>3=x#
The answer is #x=3#.
