# How do you simplify (1/8) ^ (5/2)?

${\left(\frac{1}{8}\right)}^{\frac{5}{2}} = {\left(\frac{1}{2}\right)}^{\frac{15}{2}} = \frac{\sqrt{2}}{256}$

#### Explanation:

We can first work this and keep it in terms of exponents. Since $1 = {1}^{3} \mathmr{and} 8 = {2}^{3}$, we can write:

${\left(\frac{1}{8}\right)}^{\frac{5}{2}} = {\left({1}^{3} / {2}^{3}\right)}^{\frac{5}{2}} = {\left({\left(\frac{1}{2}\right)}^{3}\right)}^{\frac{5}{2}} = {\left(\frac{1}{2}\right)}^{3 \times \frac{5}{2}} = {\left(\frac{1}{2}\right)}^{\frac{15}{2}}$

(this also uses the rule that ${\left({x}^{a}\right)}^{b} = {x}^{a b}$)

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We can also evaluate this numerically.

Let's first see that the exponent $\frac{15}{2}$ means that we're going to take the fraction $\frac{1}{2}$ and take the square root (that's the 2 in $\frac{15}{2}$) and also take $\frac{1}{2}$ to the 15th power (that's the 15 in $\frac{15}{2}$).

Let's talk about the numerator first.

${1}^{\frac{15}{2}} = 1$

And now the denominator:

${\left(2\right)}^{\frac{15}{2}} = {2}^{\frac{14}{2} + \frac{1}{2}} = {2}^{7 + \frac{1}{2}} = {2}^{7} \times {2}^{\frac{1}{2}} = 128 \sqrt{2}$

(we used the rule that ${x}^{a} \times {x}^{b} = {x}^{a + b}$)

We can then say that:

${\left(\frac{1}{2}\right)}^{\frac{15}{2}} = \frac{1}{128 \sqrt{2}}$

We can then rationalize the denominator:

$\frac{1}{128 \sqrt{2}} \left(\frac{\sqrt{2}}{\sqrt{2}}\right) = \frac{\sqrt{2}}{128 \times 2} = \frac{\sqrt{2}}{256}$