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

##### 1 Answer
Feb 9, 2017

${\left(81 {x}^{12}\right)}^{1.25} = 243 {\left\mid x \right\mid}^{15}$

#### Explanation:

The identity:

${\left({x}^{a}\right)}^{b} = {x}^{a b}$

holds under any of the following conditions:

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

• $x = 0$ with $a , b \ge 0$.

• $x < 0$ with $a , b$ any integers.

In other circumstances it can fail.

$\textcolor{w h i t e}{}$
Given:

${\left(81 {x}^{12}\right)}^{1.25}$

FIrst note that:

${\left(81 {x}^{12}\right)}^{1.25} = {\left({3}^{4} {\left({x}^{3}\right)}^{4}\right)}^{\frac{5}{4}} = {\left({\left(3 {x}^{3}\right)}^{4}\right)}^{\frac{5}{4}}$

If $x \ge 0$ then $3 {x}^{3} \ge 0$, so since both $4 \ge 0$ and $\frac{5}{4} \ge 0$ we can assert:

${\left({\left(3 {x}^{3}\right)}^{4}\right)}^{\frac{5}{4}} = {\left(3 {x}^{3}\right)}^{4 \cdot \frac{5}{4}} = {\left(3 {x}^{3}\right)}^{5} = {3}^{5} {x}^{15} = 243 {x}^{15}$

If $x < 0$ then $3 {x}^{3} < 0$, but ${x}^{12} = {\left(- x\right)}^{12}$, so:

${\left(81 {x}^{12}\right)}^{1.25} = {\left(81 {\left(- x\right)}^{12}\right)}^{1.25} = 243 {\left(- x\right)}^{15} = - 243 {x}^{15}$

To cover both cases, we can write:

${\left(81 {x}^{12}\right)}^{1.25} = 243 {\left\mid x \right\mid}^{15}$