SOLUTION: (6x^7)^-4 ___________ = 1/3x^20 (2x^2) I keep coming up with this answer and I'm not sure if I have it right. Can you please help me?

Algebra ->  Exponents -> SOLUTION: (6x^7)^-4 ___________ = 1/3x^20 (2x^2) I keep coming up with this answer and I'm not sure if I have it right. Can you please help me?      Log On


   

Question 106262This question is from textbook
: (6x^7)^-4
___________ = 1/3x^20
(2x^2)
I keep coming up with this answer and I'm not sure if I have it right. Can you please help me? This question is from textbook

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Exponentiation rules:
%28XY%29%5EM=X%5EM%2AY%5EM
X%5EM%2AX%5EN=X%5E%28M%2BN%29
X%5EM%2FX%5EN=X%5E%28M-N%29
Let's work on the numerator first and simplify.
NUM:%286x%5E7%29%5E-4=6%5E%28-4%29x%5E%287%2A-4%29
NUM:%286x%5E7%29%5E-4=6%5E%28-4%29x%5E%28-28%29
Therefore,
%286x%5E7%29%5E%28-4%29%2F%282x%5E2%29=6%5E%28-4%29x%5E%28-28%29%2F%282x%5E2%29
%286x%5E7%29%5E%28-4%29%2F%282x%5E2%29=6%5E%28-4%29%2A2%5E%28-1%29%2Ax%5E%28-28-2%29
%286x%5E7%29%5E%28-4%29%2F%282x%5E2%29=6%5E%28-4%29%2A2%5E%28-1%29%2Ax%5E%28-30%29
%286x%5E7%29%5E%28-4%29%2F%282x%5E2%29=1%2F%282592%2Ax%5E%2830%29%29