Answer: 
Step-by-step explanation:
By the negative exponent rule, you have that:

By the exponents properties, you know that:

Therefore, you can rewrite the left side of the equation has following:

Descompose 32 and 8 into its prime factors:

Rewrite:

Then:

As the base are equal, then:

Solve for x:

Answer:
(4-1)•n=n•(4-1)
Step-by-step explanation:
It should equal each other so no matter what n equals it will equal each other :3
C
If you plug in the values in that equation, it works.
Answer:
no, this is not a function
Step-by-step explanation:
i hope this helps :)