How do you take a variable with a negative or non negative exponent, and then negatively square it, cube it, or whatever?
For example, what would (y^5)^-5 or (y^-5)^-5 look like?
Exponents question?
2008-05-03 9:41 am
回答 (8)
2008-05-03 9:59 am
✔ 最佳答案
Paul is wrong.When raising power to another power, you multiply the exponenents.
so (y^5)^-5 is the same as
1/ (y^5)^5 which = 1/y^25
same with (y^-5)^-5, that would equal
1/(y^-5)^5 is the same as
1/ (1/y^5)^5 which =
(y^5)^5 which then equals = y^25
2008-05-03 5:07 pm
1)
(y^5)^-5
= y^[(5)(-5)]
= y^-25
2)
(y^-5)^-5
= y^[(-5)(-5)]
= y^25
(y^5)^-5
= y^[(5)(-5)]
= y^-25
2)
(y^-5)^-5
= y^[(-5)(-5)]
= y^25
2008-05-03 5:05 pm
(y^5)^-5 = y^(--25) or (y^-5)^-5 = y^25
2008-05-03 4:57 pm
Remember these laws on exponents:
(a^m)^n = a^m*n
example: (a^2)^3 = a^2*3 = a^6
a^-m = 1/(a^m)
example: a^-2 = 1/(a^2)
Therefore:
(y^5)^-5 = y^(5*-5) = y^-25 = 1/(y^25)
(y^-5)^-5 = y^(-5*-5) = y^25
(a^m)^n = a^m*n
example: (a^2)^3 = a^2*3 = a^6
a^-m = 1/(a^m)
example: a^-2 = 1/(a^2)
Therefore:
(y^5)^-5 = y^(5*-5) = y^-25 = 1/(y^25)
(y^-5)^-5 = y^(-5*-5) = y^25
2008-05-03 4:56 pm
(x^a)^b = x^(ab)
(y^5)^-5 = y^(5Ã(-5)) = y^-25
(y^-5)^-5 = y^((-5)(-5)) = y^25
(y^5)^-5 = y^(5Ã(-5)) = y^-25
(y^-5)^-5 = y^((-5)(-5)) = y^25
2008-05-03 4:55 pm
x^(-n) = 1/x^n
So:
(y^5)^-5 = y ^(-25) = 1/ y^25
(y^-5)^-5= y^(-5*-5) = y^25
So:
(y^5)^-5 = y ^(-25) = 1/ y^25
(y^-5)^-5= y^(-5*-5) = y^25
2008-05-03 4:55 pm
(y^a)^(b) = y^(ab)
(y^5)^(-5) = y^(-25)
y^(-5)^(-5) = y^25
(y^5)^(-5) = y^(-25)
y^(-5)^(-5) = y^25
2008-05-03 4:52 pm
(y^5)^-5 = y^(5-5) = y^0 = 1
(y^-5)^-5 = y^(-5-5) = y^-10 = 1/y^10
(y^-5)^-5 = y^(-5-5) = y^-10 = 1/y^10
收錄日期: 2021-05-01 10:26:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080503014122AAnUq8j