# How do you simplify (1/2k^8v^3)^2(60kv^4)?

Jun 24, 2018

$15 {k}^{17} {v}^{10}$

#### Explanation:

${\left(\frac{1}{2} {k}^{8} {v}^{3}\right)}^{2} \left(60 k {v}^{4}\right) =$
$\frac{1}{4} {k}^{16} {v}^{6} \cdot 60 k {v}^{4} =$
$15 {k}^{17} {v}^{10}$

Jun 24, 2018

To get our answer, $15 {k}^{17} {v}^{10}$, we can use some exponent rules for simplification.

#### Explanation:

Let's first simplify the left side of the expression, ${\left(\frac{1}{2} {k}^{8} {v}^{3}\right)}^{2}$. Notice how there is a coefficient and two variables. Each one will be squared when simplified. Let's also keep in mind that ${\left({a}^{b}\right)}^{c} = {a}^{b \cdot c}$:

${\left(\frac{1}{2}\right)}^{2} = {1}^{2} / {2}^{2} = \frac{1}{4}$
${\left({k}^{8}\right)}^{2} = {k}^{8 \cdot 2} = {k}^{16}$
${\left({v}^{3}\right)}^{2} = {v}^{3 \cdot 2} = {v}^{6}$

We now have $\frac{1}{4} {k}^{16} {v}^{6}$. Let's multiply that by the other half of the expression. Remember that ${a}^{b} \cdot {a}^{c} = {a}^{b + c}$:

$\frac{1}{4} \cdot 60 = 15$
${k}^{16} \cdot k = {k}^{16 + 1} = {k}^{17}$
${v}^{6} \cdot {v}^{4} = {v}^{6 + 4} = {v}^{10}$

We now have our final expression, $15 {k}^{17} {v}^{10}$.