Indices and double negatives

Simplify `(x^-4)^-2`

To tackle this problem try simple numbers you know first.

We know

`10^-2=1/10^2=1/(10\times10)=1/100=0.01`

Therefore `x^(-4)=1/(x^4)`

And therefore `(x^-4)^-2=(1/x^4)^-2`

Which equals `1/((1/(x^4))^2`

And `1/{1/x^4\times1/x^4`

And as `1/2\times1/2=1/4`

Then `1/{1/x^4\times1/x^4\}=1/{1/x^8}`

power to 16 image

`x^8/1`

We get `x^8`

Answer:

`(x^-4)^-2=\x^8`

This makes sense because two negatives make a positive, but it’s excellent to check.