Question

I need to evaluate log base 3 (1/162)

Answer 1

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
$(log_3 \frac{1}{2})-4$