Question

How do you evaluate `3^ (3log_3 2+ log_3 8)`?

Answer 1

I found: `64`

Explanation:

The sum of the exponents allow us to write:
`3^(3log_3(2))*3^(log_3(8))=`
we can now use the property of the logs and definitiom to write:
`=3^(log_3(2^3))*3^(log_3(8))=`
and definition of log:
`=2^3*8=64`