How do you multiply #3y x ^ { - 3} \cdot x ^ { 4} y ^ { 0}#?

1 Answer
anor277
Nov 17, 2016

#3yx^(-3)xxx^4y^0# #=# #3xy#

Explanation:

#3yx^(-3)xxx^4y^0# #=# #3y*y^0xxx^(-3)xxx^4#

Now when we mulitply #x^axxx^b#, we simply sum the exponents:

#x^axxx^b# #=# #x^(a+b)#

Thus #x^(-3)xxx^4=x^(-3+4)=x^1=x#,

And #3yxxy^0=3xxy^(1+0)=3y^1=3y#

And thus #3yx^(-3)xxx^4y^0# #=# #3xy#

Of course, you realize that something raised to the power of #0=1#, i.e. #x^0=y^0=10^0, etc.=1#.

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