Question

FIND THE MISSING EXPONENT
x^ • x^3 = 1/x^2

Answer 1

Explanation:

Hi Ashley, this looks harder than it really is :)

recall your powers rules

so
`x^(2)` *
`x^(2)` =
`x^(2+2)`=
`x^(4)`

knowing that you can probably figure this out. But I'll help anyway

`x^(?+3)` = 1 /
`x^(2)`

now flip over the fraction to make this even easier

`x^(?+3)` =
`x^(-2)`

ohhhh I bet you can see it now, huh

we just need 3 plus something to equal -2

that's negative 5 , right???

`x^(-5+3)` =
`x^(-2)`

ahh, nice , now just break it back up

`x^(-5)` *
`x^(3)` = 1 /
`x^(2)`

got it???

Answer 2

Explanation:

x^y × x^3 = 1/(x^2) = x^(-2)

that means

y + 3 = -2

y = -5

the missing exponent is -5

x^(-5) × x^3 = 1/(x^2)