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 (1)/(162)`=y

`log_3 (1)/(2*3^4)`=y

`log_3 (1)/(2) + log_3 (1)/(3^4)`=y

`log_3 (1)/(2) + log_3 (3^(-4))`=y

`log_3 (1)/(2) -4`=y
the result is

`(log_3 (1)/(2))-4`