#1#. 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 simplify the left side of the equation.
#log(3x+7)+log(x-2)=1#
#log((3x+7)(x-2))=1#
#log(3x^2+x-14)=1#
#2#. 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(3x^2+x-14)=log(10)#
#3#. 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.
#3x^2+x-14=10#
#4#. Subtract #10# from both sides of the equation.
#3x^2+x-24=0#
#5#. Factor the quadratic equation.
#(3x-8)(x+3)=0#
#6#. Set each factor to #0# and solve for #x#.
#3x-8=0color(white)(XXXXXXX)x+3=0#
#3x=8color(white)(XXXXXXXXX)color(red)cancelcolor(green)(|bar(ul(color(white)(a/a)x=-3color(white)(a/a)|)))#
#color(green)(|bar(ul(color(white)(a/a)x=8/3color(white)(a/a)|)))#
#7#. However, #color(red)(x=-3)# does not satisfy the equation since it produces a log with a negative number, and you #color(teal)"can't log a negative number"#.
For example, when you substitute #color(red)(x=-3)# back into the original equation:
#log(3x+7)+log(x-2)=1#
#log(3color(red)((-3))+7)+log(color(red)((-3))-2)=1#
#log(-9+7)+log(-5)=1#
#color(teal)(log(-2))+color(teal)(log(-5))=1#
For this reason, #color(red)(x=-3)# is not a solution of the equation.
#:.#, #x=8/3#.
