Nth Roots of nth Powers

$a$	$n$	$n a^{n}$	Is $n$ even?	Simplify
$4$	$3$	$3 4^{3}$	$\times$	$4$
$- 6$	$5$	$5 (- 6)^{5}$	$\times$	$- 6$
$2$	$6$	$6 2^{6}$	$✓$	$∣ 2 ∣ = 2$
$- 3$	$4$	$4 (- 3)^{4}$	$✓$	$∣ - 3 ∣ = 3$

a

n

n a^{n}

n

even?

Simplify

4

3

3 4^{3}

\times

4

- 6

5

5 (- 6)^{5}

\times

- 6

2

6

6 2^{6}

✓

∣ 2 ∣ = 2

- 3

4

4 (- 3)^{4}

✓

∣ - 3 ∣ = 3

To write the expression for $n a^{n},$ there are two cases to consider.

$a \geq 0$
$a < 0$

Both cases will be considered one at a time.

$a \geq 0$

Start by noting that since

a

is non-negative,

a^{n}

is also non-negative. This means that the

n^{th}

root can be rewritten using a rational exponent.

n a^{n} = (a^{n})^{\frac{1}{n}}

Since both

n

and

\frac{1}{n}

are rational numbers, the Power of a Power Property for Rational Exponents can be applied to simplify the obtained expression.

(a^{n})^{\frac{1}{n}}

▼

Simplify

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

a^{n \cdot \frac{1}{n}}

$a \cdot \frac{1}{a} = 1$

a^{1}

ExponentOne

$a^{1} = a$

a

Therefore,

n a^{n}

is equal to

a

a

is non-negative.

$a < 0$

In case of a negative value of $a,$ there are also two cases two consider.

$n$ is even
$n$ is odd

Even $n$

Recall that a root with an even index

n_{e}

is defined only for non-negative numbers. Although

a

is negative,

a^{n_{e}}

is positive. Also, a power with a negative base and an even exponent can be rewritten as a power with a positive base.

a^{n_{e}} = (- a)^{n_{e}}

Now, since

- a

is positive, the Power of a Power Property for Rational Exponents can be applied again to simplify

n_{e} a^{n_{e}} .

n_{e} a^{n_{e}}

▼

Simplify

NegBaseToPosPow

$(- a)^{n_{e}} = a^{n_{e}}$

n_{e} (- a)^{n_{e}}

RootToPowD

$n a = a^{\frac{1}{n}}$

((- a)^{n_{e}})^{\frac{1}{n _{e}}}

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

(- a)^{n_{e} \cdot \frac{1}{n _{e}}}

$a \cdot \frac{1}{a} = 1$

(- a)^{1}

ExponentOne

$a^{1} = a$

- a

Odd $n$

A root with an odd index

n_{o}

is defined for all real numbers. By the definition of the

n^{th}

root, the expression

n_{o} a^{n_{o}}

is the number

y

that, when multiplied by itself

n_{o}

times, will result in

a^{n_{o}} .

n_{o} times y \cdot y \cdot \dots \cdot y = a^{n_{o}}

Because

n_{o}

is odd and

a

is negative,

a^{n_{o}}

is also negative. This means that the best candidate for

y

is simply

a .

n a^{n} = a

Summary

If $a$ is non-negative, $n a^{n}$ is always equal to $a .$ However, in case of negative $a,$ the value of $n a^{n}$ depends on the parity of $n .$

	$a \geq 0$	$a < 0$
Even $n$	$n a^{n} = a$	$n a^{n} = - a$
Odd $n$	$n a^{n} = a$	$n a^{n} = a$

To conclude, for odd values of $n,$ the expression $n a^{n}$ is equal to $a .$ On the other hand, if $n$ is even, $n a^{n}$ can be written as $∣ a ∣ .$

n a^{n} = {∣ a, ∣ a ∣, if n is odd if n is even

$a$	$m$	$n$	$n a^{m}$	Simplify
$2$	$8$	$4$	$4 2^{8}$	$2^{\frac{8}{4}} = 2^{2} = 4$
$- 3$	$6$	$2$	$(- 3)^{6}$	$∣ ∣ ∣ ∣ (- 3)^{\frac{6}{2}} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ (- 3)^{3} ∣ ∣ ∣ = ∣ - 27 ∣ = 27$
$x$	$6$	$3$	$3 x^{6}$	$x^{\frac{6}{3}} = x^{2}$
$- 3$	$2$	$8$	$8 (- 3)^{2}$	$4 ∣ - 3 ∣ = 43$
$2$	$3$	$6$	$6 2^{3}$	$2$
$- 2$	$3$	$9$	$9 (- 2)^{3}$	$3 - 2$
$2$	$6$	$8$	$8 2^{6}$	$4 2^{3} = 48$
$- 2$	$6$	$8$	$8 (- 2)^{6}$	$4 ∣ - 2 ∣^{3} = 4 2^{3} = 48$

a

m

n

n a^{m}

Simplify

2

8

4

4 2^{8}

2^{\frac{8}{4}} = 2^{2} = 4

- 3

6

2

(- 3)^{6}

∣ ∣ ∣ ∣ (- 3)^{\frac{6}{2}} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ (- 3)^{3} ∣ ∣ ∣ = ∣ - 27 ∣ = 27

x

6

3

3 x^{6}

x^{\frac{6}{3}} = x^{2}

- 3

2

8

8 (- 3)^{2}

4 ∣ - 3 ∣ = 43

2

3

6

6 2^{3}

2

- 2

3

9

9 (- 2)^{3}

3 - 2

2

6

8

8 2^{6}

4 2^{3} = 48

- 2

6

8

8 (- 2)^{6}

4 ∣ - 2 ∣^{3} = 4 2^{3} = 48

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$n^{th}$ Roots of $n^{th}$ Powers

Proof

$a \geq 0$

$a < 0$

Even $n$

Odd $n$

Summary

Extra

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Proof

a≥0

a<0

Even n

Odd n

Summary

Extra

$a \geq 0$

$a < 0$

Even $n$

Odd $n$