#1#. Bring all the logs to the left side of the equation.
#log(x+6)=1-log(x-5)#
#log(x+6)+log(x-5)=1#
#2#. Use the log property, #log_color(purple)b(color(red)m*color(blue)n)=log_color(purple)b(color(red)m)+log_color(purple)b(color(blue)n)#, to rewrite the left side of the equation.
#log((x+6)(x-5))=1#
#3#. Use the log property, #log_color(purple)b(color(purple)b^color(orange)x)=color(orange)x#, to rewrite the right side of the equation.
#log((x+6)(x-5))=log(10)#
#4#. Since the equation now follows a "#log=log#" situation, where the bases are the same on both sides, rewrite the equation without the "log" portion.
#(x+6)(x-5)=10#
#5#. Expand the brackets.
#x^2+x-30=10#
#6#. Subtract #10# from both sides.
#x^2+x-40=0#
#7#. Use the quadratic formula to solve for #x#.
#color(darkorange)(a=1)color(white)(XXXXXX)color(teal)(b=1)color(white)(XXXXXX)color(violet)(c=-40)#
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-(color(teal)1)+-sqrt((color(teal)1)^2-4(color(darkorange)1)(color(violet)(-40))))/(2(color(darkorange)1))#
#x=(-1+-sqrt(1+160))/2#
#color(green)(|bar(ul(color(white)(a/a)x=(-1+-sqrt(161))/2color(white)(a/a)|)))#
