Question 975452


{{{0= log(5, (x-2)) - log(5 ,(2x))}}}

{{{0= log(5, (x-2)/2x)}}}.......since {{{log(5,1)=0}}}, we have

{{{log(5,1)= log(5, (x-2)/2x)}}}....since log and bases same, we have

{{{1= (x-2)/2x}}}......solve for {{{x}}}

{{{1*2x= (x-2)}}}

{{{2x= x-2}}}

{{{2x-x=-2}}}

{{{x=-2}}}