# How do you simplify (9a^3b^4)^(1/2)?

Aug 30, 2016

$3 {b}^{2} {a}^{\frac{3}{2}}$ or $3 a {b}^{2} \sqrt{a}$ depending on what is meant by "simplify".

#### Explanation:

Exponents distribute across multiplication, so

${\left(9 {a}^{3} {b}^{4}\right)}^{\frac{1}{2}} = {9}^{\frac{1}{2}} {\left({a}^{3}\right)}^{\frac{1}{2}} {\left({b}^{4}\right)}^{\frac{1}{2}}$

Now use ${9}^{\frac{1}{2}} = \sqrt{9} = 3$.

Also use ${\left({x}^{a}\right)}^{b} = {x}^{a b}$, to get

${\left({a}^{3}\right)}^{\frac{1}{2}} = {a}^{\frac{3}{2}}$ which can be written $a \cdot {a}^{\frac{1}{2}} = a \sqrt{a}$

and ${\left({b}^{4}\right)}^{\frac{1}{2}} = {b}^{2}$.

${\left(9 {a}^{3} {b}^{4}\right)}^{\frac{1}{2}} = {9}^{\frac{1}{2}} {\left({a}^{3}\right)}^{\frac{1}{2}} {\left({b}^{4}\right)}^{\frac{1}{2}}$

$= 3 {a}^{\frac{3}{2}} {b}^{2}$ $\text{ }$ (which is simpler in some sense)

$= 3 a \sqrt{a} {b}^{2}$

which is not as easy to read as putting the radical last

$= 3 a {b}^{2} \sqrt{a}$.

Aug 30, 2016

$3 {a}^{\frac{3}{2}} {b}^{2}$

#### Explanation:

Using the $\textcolor{b l u e}{\text{laws of exponents}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\left({a}^{m}\right)}^{n} = {a}^{m n}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This law applies to each value inside the bracket.

rArr9^(1xx1/2xxa^(3xx1/2)xxb^(4xx1/2)=9^(1/2)a^(3/2)b^2

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{\frac{1}{2}} = \sqrt{a}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow {9}^{\frac{1}{2}} = \sqrt{9} = 3$

$\Rightarrow {\left(9 {a}^{3} {b}^{4}\right)}^{\frac{1}{2}} = 3 {a}^{\frac{3}{2}} {b}^{2}$