[tex]\frac{y^{-7^{}}}{x^{7^{}}x^{8^{}}y^{-7^{}}}[/tex]
We can cancel y^-7 up with y^-7 down
Because in the division with power we subtract the power of the same base
So 7^-1/7^-1 = 7^(-1)-(-1) = 7^(-1+1) = the 7^(0) = 1
[tex]\frac{1}{x^{7^{}}x^{8^{}}}[/tex]
Now let us add the powers of x because in multiplication we add the power of same base
x^7 * x^8 = x^(7 + 8) = x^15
The answer is 1/x^(15)
The first answer
