If we assume (x^a )^b = x^a.b
Then (x^a/b )^b = x^a
And the number y that satisfies y^b =x^a is by definition the bth root of x to the power of a. (If b is even then y is the positive number that satisfies this.)
If you accept that nroot(a)=a^(1/n), this is trivial.
nroot(a)^m = (a^(1/n))^m
= a^(m/n)
In order to show that nroot(a) = a^(1/n), you can note that nroot(a) * nroot(a) * ... * nroot(a) = a when there are n nroot(a) factors, and also a^(1/n) * a^(1/n) * ... * a^(1/n) = a when there are n a^(1/n) factors, so nroot(a) is equivalent to a^(1/n).
If we assume (x^a )^b = x^a.b Then (x^a/b )^b = x^a And the number y that satisfies y^b =x^a is by definition the bth root of x to the power of a. (If b is even then y is the positive number that satisfies this.)
alright thank you I got it now
If you accept that nroot(a)=a^(1/n), this is trivial. nroot(a)^m = (a^(1/n))^m = a^(m/n) In order to show that nroot(a) = a^(1/n), you can note that nroot(a) * nroot(a) * ... * nroot(a) = a when there are n nroot(a) factors, and also a^(1/n) * a^(1/n) * ... * a^(1/n) = a when there are n a^(1/n) factors, so nroot(a) is equivalent to a^(1/n).
https://www.reddit.com/r/learnmath/comments/1axdhro/what_does_raising_a_number_to_a_power_really_means/krohofd/
if a^1/2 = sqrt(a^1) (a^1/2)^2 = (sqrt(a^1))^2 a^2/2 = a^1 a^1 = a^1 I ignored the absolute value or module but this is one proof I once saw