log(a*b)=log(a)+log(b)
and
logₐbˣ=xlogₐb
also
x^-m=1/(x^m)
and
logₐaˣ=x
so
\(log_3 \frac{1}{162}\)=y
\(log_3 \frac{1}{2*3^4}\)=y
\(log_3 \frac{1}{2} + log_3 \frac{1}{3^4}\)=y
\(log_3 \frac{1}{2} + log_3 (3^{-4})\)=y
\(log_3 \frac{1}{2} -4\)=y
the result is
\((log_3 \frac{1}{2})-4\)
