4 years ago

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

1 answer:

5 0

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

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