Raising a Power to a PowerDate: 11/27/97 at 14:21:57 From: Lucas Overmire Subject: (Algebra) raising a power to a power I know that this answer is right but I just can't seem to figure out why. Could you please help me? [(-x)to the 5 power]to the 8 power. My teacher said that the answer was x to the 40 power but I don't understand why it isn't -x. Thanks for any help. Lucas Overmire Date: 11/27/97 at 18:21:49 From: Doctor Charles Subject: Re: (Algebra) raising a power to a power (-x) ^ 5 = (-x) * (-x) * (-x) * (-x) * (-x) ((-x)^5)^8 = ((-x)^5) * ((-x)^5) * ((-x)^5) * ((-x)^5) * ((-x)^5) * ((-x)^5) * ((-x)^5) * ((-x)^5) = (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) * (-x) = (-x) ^ 40 or = (-x * -x) ^ 20 = ( x ^ 2 ) ^ 20 = x ^ 40 If we ended up with an odd power things would have been different: for example ((-x) ^ 5) ^ 7 = (-x) ^ 35 = (-x) * ((-x) ^ 34) = (-x) * (-x * -x) ^ 17 = (-x) * ( x ^ 2) ^ 17 = (-x) * x^34 = - (x * x^34) = - x^35 In general (-x)^n = x^n if n is even and: (-x)^n = -x^n if n is odd. We can prove this. If n is odd then n = 2*m+1 for some m (-x)^n = (-x)^(2m+1) = (-x) * (-x) ^ 2m = (-x) * ((-x)^2)^m = (-x) * (x^2)^m = (-x) * x^2m = - x^(2m+1) = - x^n If n is even then n = 2m for some m so: (-x)^n = (-x) ^ 2m = ((-x)^2)^m = (x^2)^m = x^2m = x^n -Doctor Charles, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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