Answer :
Answer:
In 1st case the multiplication is to be done for the k times, in the second case the number of multiplication is given as [tex]2^k-1[/tex] times.
As the value of k will always be less than that of [tex]2^k-1[/tex], thus the case 1 is an efficient way of finding the values.
Step-by-step explanation:
Case No 1 is given as:
In this case the value of [tex]{x^2}^k[/tex] is found by subsequent squaring such that
[tex]{{x^2}^k}=x(x^2.x^4.x^8......{x^2}^k)\\{{x^2}^k}=x({x^2}^1.{x^2}^2.{x^2}^3......{x^2}^k)\\[/tex]
So in this case the multiplication is to be done for the k times.
Case No 2 is given as
[tex]{x^2}^k=x(x.x.x.x.x...x) 2^k-1 times[/tex]
So in this case the number of multiplication is given as [tex]2^k-1[/tex] times.
As the value of k will always be less than that of [tex]2^k-1[/tex], thus the case 1 is an efficient way of finding the values.