Question

How do you simplify `\frac { gh ^ { - 9} \cdot g ^ { 0} h } {gh ^ { 0} }` with positive exponents?

Answer 1

`1/h^8`

Explanation:

Remember that `a^0 = 1` for any `a ne 0`, so assuming neither `g` or `h` are zero (if either of them were, we'd be dividing by zero which isn't allowed anyways), we can say:

`(gh^(-9)*g^0h)/(gh^0) = (gh^(-9)*1*h)/(g*1) = (gh^(-9)*h)/g`

The lonely `g` can be crossed out on both numerator and denominator, so

`(gh^(-9)*h)/g = h^(-9)*h`

And the law of exponents say that `a^b*a^c = a^(b+c)`, so we can say

`h^(-9)*h = h^(-9+1) = h^(-8)`

And finally, `a^(-b) = 1/a^b`, so

`h^(-8) = 1/h^8`