by Guest » Sat Nov 12, 2011 5:26 pm
i meant to take a preview with the last post sry clicked submit by accident
...
we need to change the base to 3 after that we need to use [tex]log_{a}{bc}=log_{a}b +log_{a}c[/tex]
than the expression will have ingegers only and [tex]log_{3}2[/tex]
for the sake of simplicity lets say that [tex]log_{3}2=b[/tex]
after simplifying the expression we get [tex]\frac {6b^2+5b+1}{6b^2+5b+1}=1[/tex]