Question 1096770
{{{log(b,(2))=A}}}
{{{log(b,(3))=C}}}
Then,
{{{log(b,(sqrt(2/27)))=(1/2)log(b,(2/27))}}}
{{{log(b,(sqrt(2/27)))=(1/2)(log(b,(2))-log(g,(27)))}}}
{{{log(b,(sqrt(2/27)))=highlight((A-C)/2)}}}