Question 11901
  {{{log(x, 5) + log(5, x) = 2}}}
 
 Let {{{y  = log(5, x)}}},then we have  y + 1/y = 2,

 so {{{y^2- 2y + 1 = 0 }}}

 Hence, y = 1 and then x = {{{5^y }}} ???

 It is very staightford, I think you can understand.

 Kenny