Question 362207
{{{ log( x^2, 9 ) - log( x, 36 ) = 1 }}}

See the rule : {{{ log( a^n, b^m )  }}} = {{{ (m/n) * log(a,b) }}}

In your question,{{{ log( x^2, 3^2 )  }}} = {{{ (2/2)* log(x,3) }}}

The problem becomes;

{{{ log(x,3) - log( x, 36 ) = 1 }}}

See rule 2 :  {{{ log( a,(x/y) )  }}} = {{{ log( a,x )  }}} - {{{ log(a,y) }}} 

So that >> {{{ log(x,3) - log( x, 36 ) = 1 }}} = {{{ log(x,3/36) }}}

Therefore >> {{{ log(x,1/12) = 1 }}}

See rule 3 : {{{ log(a,a) = 1 }}}

Therefore >> x = 1/12

RF.