Question

Help with exponents!

Question posted below... MEDAL!

Answer 1

\[\huge (x^4)^{-3} \times 2x^4\]

Answer 2

@hartnn

Answer 3

first term

\(\huge (a^b)^c = a^{bc}\)

Answer 4

you subtracted the exponents,
the rule tell us to multiply them :3

Answer 5

@hartnn agree

Answer 6

Oh my godd

Answer 7

I am running our of my mind :\

Answer 8

*out

Answer 9

\(4 \times (-3) = - (4\times 3) = .. \)

Answer 10

\[\huge x^{4 \times (-3)} = x^{-12}\]

Answer 11

yussss

Answer 12

now
\(\huge a^b \times a ^c = a^{b+c}\)

Answer 13

2 \((x^{-12} \times x^4) = .. ?\)

Answer 14

So here's the thing, I don't why we have to combine them... Is it because 2x^4 has two terms: 2 and x^4?

Answer 15

theres a common base, \(x\)

Answer 16

2 is a constant, lets throw that out of the process of multiplying 'x's

Answer 17

So just to clarify, if we have like 45x^5 then i will just have to take out the 45 and work with the x being the common base with another power right?

Answer 18

....with another term of x.

yes.

Answer 19

\[\large 2(x^{-8}) = 2x^{-8}\]

Answer 20

Is that correct ?

Answer 21

\(2 x^3 \times 45 x^5 = 2\times 45 x^{3+5}\)

Answer 22

yes, thats correct :)

Answer 23

2/x^8 also "looks" nice

Answer 24

My answer is \[\huge \frac{ 2 }{ x^8 }\]

Answer 25

\(\huge \checkmark \)

Answer 26

The reason for why I have to put the x^8 to the denominator is because of the exponent of -8 right? The negative exponent tells the base to get to its reciprocal and then the exponent becomes positive?

Answer 27

correcto!

\(\Large a^{-b} = \dfrac{1}{a^b}\)

Answer 28

Thank you!

Answer 29

welcome ^_^