Question

How do you simplify `(81x^12)^1.25`?

Answer 1

`(81x^12)^1.25 = 243 abs(x)^15`

Explanation:

The identity:

`(x^a)^b = x^(ab)`

holds under any of the following conditions:

• `x > 0` with `a, b` any real numbers.

• `x = 0` with `a, b >= 0`.

• `x < 0` with `a, b` any integers.

In other circumstances it can fail.

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Given:

`(81x^12)^1.25`

FIrst note that:

`(81x^12)^1.25 = (3^4 (x^3)^4)^(5/4) = ((3x^3)^4)^(5/4)`

If `x >= 0` then `3x^3 >= 0`, so since both `4 >= 0` and `5/4 >= 0` we can assert:

`((3x^3)^4)^(5/4) = (3x^3)^(4*5/4) = (3x^3)^5 = 3^5 x^15 = 243 x^15`

If `x < 0` then `3x^3 < 0`, but `x^12 = (-x)^12`, so:

`(81x^12)^1.25 = (81(-x)^12)^1.25 = 243 (-x)^15 = -243 x^15`

To cover both cases, we can write:

`(81x^12)^1.25 = 243 abs(x)^15`