Question

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

Answer 1

`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`