# How do you multiply 2w ^ { 9} u ^ { 7} \cdot 3w \cdot 2u ^ { 7}?

Jul 15, 2017

#### Answer:

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\left(2 \cdot 3 \cdot 2\right) \left({w}^{9} \cdot w\right) \left({u}^{7} \cdot {u}^{7}\right) \implies$

$12 \left({w}^{9} \cdot w\right) \left({u}^{7} \cdot {u}^{7}\right)$

Now, use these rules of exponents to complete the multiplication of the $w$ and $u$ terms:

$a = {a}^{\textcolor{b l u e}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$12 \left({w}^{9} \cdot w\right) \left({u}^{7} \cdot {u}^{7}\right) \implies$

$12 \left({w}^{\textcolor{red}{9}} \cdot {w}^{\textcolor{b l u e}{1}}\right) \left({u}^{\textcolor{red}{7}} \cdot {u}^{\textcolor{b l u e}{7}}\right) \implies$

$12 {w}^{\textcolor{red}{9} + \textcolor{b l u e}{1}} {u}^{\textcolor{red}{7} + \textcolor{b l u e}{7}} \implies$

$12 {w}^{10} {u}^{14}$

Jul 15, 2017

#### Answer:

$12 {u}^{14} {w}^{10}$

#### Explanation:

Just imagine yourself multiplying $3 w$ with $2 {w}^{9} {u}^{7}$ and $2 {u}^{7}$ with $2 {w}^{9} {u}^{7}$, then multiplying these together

First let's do the w.

$2 {w}^{9} \cdot 3 {w}^{1}$ (since $3 w = 3 {w}^{1}$)

To multiply powers, you apply the following rule

${a}^{x} \cdot {a}^{y} = {a}^{x + y}$

$\therefore 2 {w}^{9} \cdot 3 {w}^{1} = \left(2 \cdot 3\right) {w}^{9 + 1}$
$= 6 {w}^{10}$

Same applies to the u.

${u}^{7} \cdot 2 {u}^{7} = 2 {u}^{7 + 7}$
$= 2 {u}^{14}$

Now obviously, $6 \cdot 2 = 12$

Combining them, we get $12 {u}^{14} {w}^{10}$ ($u$ first because it has to be in alphabetical order)

Jul 15, 2017

#### Answer:

$6 {u}^{14} {w}^{10}$

#### Explanation:

The indices or exponents of a number of variable indicates how many of them are multiplied together.

${x}^{4} = x \cdot x \cdot x \cdot x \text{ } \leftarrow$ remember $\cdot$ means $\times$

When we multiply in algebra we multiply all the same types of bases together.

$2 \cdot {w}^{9} {u}^{7} \cdot 3 w \cdot 2 {u}^{7}$ can be re-arranged and written as:

$= \textcolor{b l u e}{2 \cdot 3} \cdot \textcolor{p u r p \le}{{u}^{7} \cdot {u}^{7}} \cdot \textcolor{f \mathmr{and} e s t g r e e n}{{w}^{9} \cdot {w}^{1}}$

$= \textcolor{b l u e}{6} \cdot \textcolor{p u r p \le}{{u}^{14}} \cdot \textcolor{f \mathmr{and} e s t g r e e n}{{w}^{10}}$

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Recall:

Note that:

$\textcolor{p u r p \le}{{u}^{7} \cdot {u}^{7} = u \cdot u \cdot u \cdot u \cdot u \cdot u \cdot u \times u \cdot u \cdot u \cdot u \cdot u \cdot u \cdot u}$

$\textcolor{p u r p \le}{{u}^{7} \cdot {u}^{7} = {u}^{14}} \text{ } \leftarrow$ add the indices of like bases

color(forestgreen)(w^9*w^1 = w*w*w*w*w*w*w*w*wxxw = w^10

color(blue)(2*3=6" "larr" "2 and 3 are just numbers, multiply as usual.