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

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Answer 1 · 2017-02-09T07:46:52

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

The identity:

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

holds under any of the following conditions:

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:

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

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