# How do you simplify (2v)^2*2v^2 and write it using only positive exponents?

Feb 8, 2017

See the entire simplification process below:

#### Explanation:

First, we will use these two rules for exponents to simplify the term on the left of this expression:

$a = {a}^{\textcolor{red}{1}}$

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${\left(2 v\right)}^{2} \cdot 2 {v}^{2} \to {\left({2}^{\textcolor{red}{1}} {v}^{\textcolor{red}{1}}\right)}^{\textcolor{b l u e}{2}} \cdot 2 {v}^{2} \to {2}^{\textcolor{red}{1} \times \textcolor{b l u e}{2}} {v}^{\textcolor{red}{1} \times \textcolor{b l u e}{2}} \cdot 2 {v}^{2} \to {2}^{2} {v}^{2} \cdot 2 {v}^{2}$

We can now use these two rule for exponents to complete the simplification:

$a = {a}^{\textcolor{red}{1}}$

${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

${2}^{\textcolor{red}{2}} {v}^{\textcolor{red}{2}} \cdot {2}^{\textcolor{b l u e}{1}} {v}^{\textcolor{b l u e}{2}} \to {2}^{\textcolor{red}{2} + \textcolor{b l u e}{1}} {v}^{\textcolor{red}{2} + \textcolor{b l u e}{2}} \to {2}^{3} {v}^{4} \to 8 {v}^{4}$