Question

How do you evaluate `log_(3) 2 times log_(2) 27`?

Answer 1

I found `3`

Explanation:

We can try changing base of the second to write:
`log_3(2)xx(log_3(27))/(log_3(2))=`
`=cancel(log_3(2))xx(log_3(27))/cancel((log_3(2)))=log_3(27)=3`