Here we have a general rule for exponents.
We will find that:
[tex](x^a)^b = x^{a*b}[/tex]
To get this, we can first check it with some values for the exponents.
a = 2
b = 3
[tex](x^2)^3 = (x^2)*(x^2)*(x^2) = (x*x)*(x*x)*(x*x) = x^6 = x^{2*3}[/tex]
Similarly, we can prove it for two general values of a and b:
[tex](x^a)^b = (x^a)*(x^a)*...*(x^a)\\b\ times[/tex]
Now each one of these parentheses has a multiplying by itself a times.
So when we write all the x's, we will see that there are a total of b*a x's
Thus we will have:
[tex](x^a)^b = (x^a)*(x^a)*...*(x^a) = x^{a*b}[/tex]
If you want to learn more, you can read:
https://brainly.com/question/5497425
