Here is how I normally solve exponents with negative numbers:
4^-2 = 4^2 = 16 = 1/16
Is that correct?
Because now I have this question I have to deal with:
2(4)^-2
So I tried solving it the same way, getting 1/16 and multiplying both of the numbers by 2 but apparently that's wrong and the answer is actually 1/8? How would I get to the answer then?
Thanks!
?
I take delivery of wolfram alpha will interpret -4^2 as -(4^2) however in case you typed four into a calculator and hit the +/- button then hit the x^2 button you'll get the postive influence. I feel those who suppose -4^2 = -16 are considering of the - signal as -1 * (number) and applying PEDMAS or BIDMAS priorities in the squaring the number first then multiplying the outcomes via -1. Whereas I see a bad number as a number in its own proper so i'd interpret -4^2 as x^2 where x = -four. This is additionally backed up through the fact that there are two solutions to x^2 = n specifically ±√n. I might be grateful if anybody with more maths than me be aware of of an authentic ruling someplace where this ambiguity is cleared up and might cite a significant source however except i'm exposed to such authorititive ruling i'll go together with what makes most feel to me and that is a terrible quantity will not be -1 * constructive number but that a poor number is in its possess proper an element of the actual number set, for this reason -2^2 = four and not -four.
Asem
2 * 4^-2
2 * 1/(4^2)
2 * 1/16
which is: 2/16
simplifying: (2/2) / (16/2)
so the answer is
1/8
iceman
4^2 = 16
4^(-2) = 1/4^2 = 1/16
Edit: How can 4^(-2) = 4^2 & 16 = 1/16, that is completely wrong people! pay attention to what you write please.
Anonymous
1/8 is right
RobertMathmanJones
Your method IS CORRECT.
But when you multiply a fraction by 2 you only multiply the NUMERATOR by 2
NOT both the numeratoe AND denominator.
Mathcerely,
Robert Jones.......Master in Mathematics (Un of Notre Dame)
"Teacher of Fine Students"
I tutor the struggling to improve grades and learning.
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