Question

How can I simplify `((f^-16)/(256g^4h^-4))^(-1/4)`?

Answer 1

`(4gf^2)/h`

Very slightly different approach.

Explanation:

Given: `((f^(-16))/(256g^4h^(-4)))^(-1/4)`

The trick is to remember that anything raised to a negative power is inverted

Rewrite as:

`((h^4)/(256g^4f^16))^(-1/4)`

Rewrite as:

`((256g^4f^16)/h^4)^(1/4)`

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For method think of: `sqrt(2^2xx2^2) = 4 = sqrt(2^2) xx sqrt(2^2)`
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Giving: `(root4(256)xxroot4(g^4)xxroot4(f^16))/(root4(h^4))`

`(4gf^2)/h`

Answer 2

`((f^(-16))/(256g^4h^(-4)))^(-1/4) = abs((4f^4g)/h)`

Explanation:

If `a, b, c > 0` then `(a^b)^c = a^(bc)`

Also `a^(-b) = 1/(a^b)`

Hence we find:

`((f^(-16))/(256g^4h^(-4)))^(-1/4)=((256g^4h^(-4))/(f^-16))^(1/4)=((4^4f^16g^4)/(h^4))^(1/4)`

`=((4^4(f^4)^4g^4)/(h^4))^(1/4)=(((4f^4g)/h)^4)^(1/4)=abs((4f^4g)/h)`

If `a > 0` then `a^(1/4)` denotes the Real positive fourth root.

Hence for any `a in RR` we have `root(4)(a^4) = abs(a)`