Sum of the roots of the equation  9log3(log2x)=log2x−(log2x)2+1  is:

9log3(log2x)=log2x−(log2x)2+1 ⇒    3log3(log2x)2=log2x−(log2x)2+1                       ⇒    (log2x)2=log2x−(log2x)2+1                          (log2x>0) ⇒    2(log2x)2−log2x−1=0 ⇒    log2x=−1/2     or  log2x=1 As log2x>0,  log2x=1    or x=2