How do you simplify (7^2/3)^(5/2)?

Apr 15, 2018

${\left(\frac{7}{\sqrt{3}}\right)}^{5}$

Explanation:

First, let's set out the exponent rules that are relevant:

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

(2) ${\left(e \cdot f\right)}^{g} = {e}^{g} \cdot {f}^{g}$

(3) ${\left(\frac{1}{h}\right)}^{j} = {h}^{-} j$

Now we first use rule (2):

${\left({7}^{2} / 3\right)}^{\frac{5}{2}} = {\left({7}^{2}\right)}^{\frac{5}{2}} \cdot {\left(\frac{1}{3}\right)}^{\frac{5}{2}}$

We can now use rule (3):

${\left({7}^{2}\right)}^{\frac{5}{2}} \cdot {\left(\frac{1}{3}\right)}^{\frac{5}{2}} = {\left({7}^{2}\right)}^{\frac{5}{2}} \cdot {3}^{- \frac{5}{2}}$

We use rule (1) to multiply out the left hand term:

${\left({7}^{2}\right)}^{\frac{5}{2}} \cdot {3}^{- \frac{5}{2}} = {7}^{2 \cdot \frac{5}{2}} \cdot {3}^{- \frac{5}{2}}$

$= {7}^{5} \cdot {3}^{- \frac{5}{2}}$

To simplify further we could again use rule (1) and rule (3):

${7}^{5} \cdot {3}^{- \frac{5}{2}} = {\left(\frac{7}{3} ^ \left(\frac{1}{2}\right)\right)}^{5}$

As ${x}^{\frac{1}{2}} \equiv \sqrt{x}$

${\left(\frac{7}{3} ^ \left(\frac{1}{2}\right)\right)}^{5} = {\left(\frac{7}{\sqrt{3}}\right)}^{5}$