Question 242934
{{{log(16,(1/2))}}} Start with the given expression.



{{{log(10,(1/2))/log(10,(16))}}} Use the change of base formula.



{{{log(10,(2^(-1)))/log(10,(16))}}} Rewrite {{{1/2}}} as {{{2^(-1)}}}



{{{log(10,(2^(-1)))/log(10,(2^4))}}} Rewrite {{{16}}} as {{{2^4}}}



{{{(-1*log(10,(2)))/(4*log(10,(2)))}}} Pull down the exponents using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}



{{{(-1*cross(log(10,(2))))/(4*cross(log(10,(2))))}}} Cancel out the common terms.



{{{-1/4}}} Simplify



So {{{log(16,(1/2))=-1/4}}}