Question 4752
Let {{{log (2,5) = x}}}

#1.
Then {{{log (2, 20) = log (2, 2^2*5)}}}

= {{{log  (2, 2^2) + log (2, 5) }}}  Notice that the first is the exponent, which is 2, and the second is x.
= {{{ 2 + x}}}


#2.  {{{log (2, (5^(1/3))/2 ) }}}

={{{log (2, (5^(1/3))) - log (2, 2^1)}}} 
= {{{(1/3) *log (2, 5) - 1}}}
= {{{(1/3) *x -1 }}}


R^2 at SCC