Question 174880
{{{2*log(3,(z))-log(3,(z-2))=2}}} Start with the given equation. 



Take note that {{{z>0}}} (for the first log) and {{{z-2>0}}}---> {{{z>2}}} (for the second log). 



So the domain is {{{z>2}}} (this interval works for BOTH logs)



{{{log(3,(z^2))-log(3,(z-2))=2}}} Rewrite the first log using the identity  {{{y*log(b,(x))=log(b,(x^y))}}}



{{{log(3,((z^2)/(z-2)))=2}}} Combine the logs using the identity {{{log(b,(A))-log(b,(B))=log(b,(A/B))}}}



{{{3^2=(z^2)/(z-2)}}} Rewrite the equation using the property: {{{log(b,(x))=y}}} ====> {{{b^y=x}}}



{{{9=(z^2)/(z-2)}}} Square 3 to get 9



{{{9(z-2)=z^2}}} Multiply both sides by {{{z-2}}}.



{{{9z-18=z^2}}} Distribute.



{{{-z^2+9z-18=0}}} Subtract {{{z^2}}} from both sides.



Notice we have a quadratic equation in the form of {{{az^2+bz+c}}} where {{{a=-1}}}, {{{b=9}}}, and {{{c=-18}}}



Let's use the quadratic formula to solve for z



{{{z = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{z = (-(9) +- sqrt( (9)^2-4(-1)(-18) ))/(2(-1))}}} Plug in  {{{a=-1}}}, {{{b=9}}}, and {{{c=-18}}}



{{{z = (-9 +- sqrt( 81-4(-1)(-18) ))/(2(-1))}}} Square {{{9}}} to get {{{81}}}. 



{{{z = (-9 +- sqrt( 81-72 ))/(2(-1))}}} Multiply {{{4(-1)(-18)}}} to get {{{72}}}



{{{z = (-9 +- sqrt( 9 ))/(2(-1))}}} Subtract {{{72}}} from {{{81}}} to get {{{9}}}



{{{z = (-9 +- sqrt( 9 ))/(-2)}}} Multiply {{{2}}} and {{{-1}}} to get {{{-2}}}. 



{{{z = (-9 +- 3)/(-2)}}} Take the square root of {{{9}}} to get {{{3}}}. 



{{{z = (-9 + 3)/(-2)}}} or {{{z = (-9 - 3)/(-2)}}} Break up the expression. 



{{{z = (-6)/(-2)}}} or {{{z =  (-12)/(-2)}}} Combine like terms. 



{{{z = 3}}} or {{{z = 6}}} Simplify. 




===================================================


Answer:



So the solutions are {{{z = 3}}} or {{{z = 6}}} 

  
  
Take note that both of these solutions are within the domain {{{z>2}}}. So they are valid solutions.