Question 1017733
1. {{{log(81,9)=log(81,sqrt(81))=log(81,(81^0.5))=0.5}}}


2. Maybe what is meant is
{{{log(8,128)=7/3}}}<-->{{{8^"7/3"=128}}}
{{{128=2*64=2*8^2=2*(2^3)^2=2*2^6=2^7}}} so {{{8^"7/3"=8^("(1/3)"*7)=(8^"1/3")^7=2^7}}}