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.