Is 1 over a^-7 the same as a^7, I think it is I just want to make sure
2007-05-21 07:03:25 · 10 answers · asked by delphiniums 1 in Science & Mathematics ➔ Mathematics
hell ya it is
2007-05-21 07:05:50 · answer #1 · answered by ? 5 · 0⤊ 0⤋
Is 1 over a^-7 the same as a^7, I think it is I just want to make sure
In exponents
(1) a^0 = 1
(2) a^1 = a
(3) (a^m)(a^n) = a^(m+n)
(4) (a^m)^n) = a^(mn)
By 1 over (a^(-7)) you mean the reciprocal of a^(-7)
By definition, the reciprocal of a number is that value by which you multiply the number to get 1. That means, that if (a)(b)=1, then a and b are reciprocals.
Let’s look at 1 and 3 above.
(a^m)(a^n) = a^(m+n).
If m = -7, what do you need n to be so that you have a^0?
You want m+n=0, m = -7
n = 7
So the reciprocal of a^(-7) = a^7.
So “1/[a^(-7)]” = a^7
2007-05-21 14:33:22 · answer #2 · answered by gugliamo00 7 · 0⤊ 0⤋
No 1 over 7 is one seventh (1/7). 7 over one is 7 (7/1)
2007-05-21 14:11:48 · answer #3 · answered by bharpman 2 · 0⤊ 2⤋
(1/a^-7) = 1/1 ⦠a^7/1 = a^7/1 = a^7
So:
Yes.
2007-05-21 14:13:04 · answer #4 · answered by Mr. Good Answers 2 · 0⤊ 0⤋
yes 1 over a^-7 = a^7
since we know that a^-m=1 over a^m
Here m=-7
2007-05-21 14:19:16 · answer #5 · answered by sriram t 3 · 0⤊ 0⤋
The rule :
a^(-b)= 1/(a^b) & vise versa
So they R the same
2007-05-21 14:20:08 · answer #6 · answered by a_ebnlhaitham 6 · 0⤊ 0⤋
yes
2007-05-21 14:06:08 · answer #7 · answered by emp211 3 · 0⤊ 0⤋
Yes indeed.
2007-05-21 14:13:26 · answer #8 · answered by ironduke8159 7 · 0⤊ 0⤋
yup
2007-05-21 14:06:00 · answer #9 · answered by sxcwhitegrl 2 · 0⤊ 0⤋
yes.
2007-05-21 14:06:27 · answer #10 · answered by Anonymous · 0⤊ 0⤋