How do you evaluate #log_[3]sqrt(18) + log_[3]sqrt(24) - log_[3](12)#?

1 Answer
Apr 9, 2016

Answer:

#log_3 sqrt(18)+log_3 sqrt(24)-log_2 (12) = 1/2#

Explanation:

In general:
#color(white)("XXX")log_b p + log_b q = log_b (pq)#
and
#color(white)("XXX")log_b r - log_b s = log_b (r/s)#

So
#color(white)("XXX")log_3 sqrt(18)+log_2 sqrt(24)-log_3(12)#

#color(white)("XXX")=log_3 ((sqrt(18)*sqrt(24))/12)#

#color(white)("XXX")= log_3 sqrt(3)#

#log_3 sqrt(3) = k# means #3^k = sqrt(3)#
#color(white)("XXX")rarr k=1/2#

So
#color(white)("XXX")log_3 sqrt(18)+log_2 sqrt(24)-log_3(12)=log_3 sqrt(3) = 1/2#