SOLUTION: negative 4z to the 6th y to the 3rd over negative 16z to the 7th y to the negative 2 and all to the negative 3rd

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: negative 4z to the 6th y to the 3rd over negative 16z to the 7th y to the negative 2 and all to the negative 3rd      Log On


   

Question 221734: negative 4z to the 6th y to the 3rd over
negative 16z to the 7th y to the negative 2 and all to the negative 3rd
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%28%28-4z%5E6y%5E3%29%2F%28-16z%5E7y%5E%28-2%29%29%29%5E%28-3%29
NOTE: If this is not your problem, please repost it with parentheses around
  • each numerator
  • each denominator
  • each expression to which an exponent applies

Since there is only one term in both the numerator and denominator and since %28a%2Ab%2Ac%29%2F%28d%2Ae%2Af%29+=+%28a%2Fd%29%2A%28b%2Fe%29%2A%28c%2Ff%29, we can simplify the fraction factor by factor:

Reducing the first fraction and using the rule for division with exponents: %28x%5Ea%29%2F%28x%5Eb%29+=+x%5E%28a-b%29 on the other two we get:
%28%281%2F4%29%2Az%5E%28-1%29%2Ay%5E5%29%5E%28-3%29
Another rule of exponents (which I call the "pseudo" (or fake) distributive property) is: %28a%2Ab%29%5Ec+=+a%5Ec%2Ab%5Ec allows us to "distribute" the exponent to each factor. (IMPORTANT: This fake distributive property does not apply if there is more than one term in the parentheses. For example %28x%2By%29%5En+%3C%3E+x%5En+%2B+y%5En!!!)
%281%2F4%29%5E%28-3%29%2A%28z%5E%28-1%29%29%5E%28-3%29%2A%28y%5E5%29%5E%28-3%29
On the first fraction we'll use the property: x%5E%28-n%29+=+1%2Fx%5En.
%284%2F1%29%5E%283%29%2A%28z%5E%28-1%29%29%5E%28-3%29%2A%28y%5E5%29%5E%28-3%29
We can cube 4 and use another rule for exponents (power of a power) on the other two: %28x%5Ea%29%5Eb+=+x%5E%28a%2Ab%29
64z%5E3y%5E%28-15%29