Question 864939
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Hi
Re TY;  Good Catch and Good work!
{{{Log(2,((x+2)^2)/(7x+2)) = log(x,(x^2))-log(sqrt(3),3) = 2-2 = 0}}}
{{{Log(2,((x+2)^2)/(7x+2)) = 0}}}
{{{(x+2)^2/(7x+2) = 1}}}  
 x^2 + 4x + 4 = 7x + 2
x^2 - 3x - 2 = 0
(x-2)(x-1) = 0,   x = 1, 2
and IF...:)
{{{log(x,(x^2)) = 2}}} and {{{log(3,3) = 1}}}
Therefore:
{{{Log(2,((x+2)^2)/(7x+2)) = 2-1 = 1}}}
{{{(x+2)^2/(7x+2) = 2}}}     |Note:  If log(2,X) = 1  then 2^1 = X
  x^2 + 4x + 4 = 14x + 4
  x^2 - 10x = 0
 x(x-10) = 0,   x = 10    |Note:  root x = 0 is an extraneous root