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

1 Answer
Jun 11, 2016

Answer:

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#