SOLUTION: Rewrite as radical and evalute: 81^3/4 Express in terms of i: a) square root of -54 b) i^53 Any help will be appreciated. thank you very much.

Algebra ->  Radicals -> SOLUTION: Rewrite as radical and evalute: 81^3/4 Express in terms of i: a) square root of -54 b) i^53 Any help will be appreciated. thank you very much.      Log On


   



Question 294383: Rewrite as radical and evalute: 81^3/4

Express in terms of i:
a) square root of -54
b) i^53

Any help will be appreciated. thank you very much.

Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
"1. Rewrite as radical and evalute: 81^3/4
2. Express in terms of i:
a) square root of -54
b) i^53
Any help will be appreciated. thank you very much."
1. rule: (x^m)(x^n) = x^(m+n)
rule: (x^m)^n = x^(mn)
+81%5E%283%2F4%29+=+81%5E%281%2F2%29+%2A+81%5E%281%2F4%29+
+81%5E%283%2F4%29+=+sqrt%2881%29+%2A+%2881%5E%281%2F2%29%29%5E%281%2F2%29+
+81%5E%283%2F4%29+=+9+%2A+%28sqrt%2881%29%5E%281%2F2%29%29+
+81%5E%283%2F4%29+=+9+%2A+9%5E%281%2F2%29+
+81%5E%283%2F4%29+=+9+%2A+sqrt%289%29+
+81%5E%283%2F4%29+=+9+%2A+3+
+81%5E%283%2F4%29+=+27+
2a. square root of -54
i^2 = -1, where +- i is the square root of -1
a^2 = -54, where a is the square root
a^2 = -1 * 54
a^2 = i^2 * (2 * 3 * 9)
a^2 = i^2 * 3^2 * 6
+a+=+3sqrt%286%29i+ or
+a+=+-3sqrt%286%29i+
2b. i^53
i^0 = 1
i^1 = i = sqrt -1
i^2 = -1
i^3 = -i
i^4 = -i * i = 1
notice this is mod 4
53 mod 4 = 1
i^53 = i^1 = i