Question 150035
demonstration is the most straight forward


using rules for exponents, (x^a)/(x^b)=x^(a-b) __ (x^5)/(x^3)=x^2 __ (x*x*x*x*x)/(x*x*x)=x*x


suppose a=b __ (x^a)/(x^b)=x^(a-b)=x^0 __ (x*x*x)/(x*x*x)=1=x^0