Write the Fraction in Simplest Form ((3z^-4)^-2)/(2v^-2z^-4)

$\frac{{(3 z^{- 4})}^{- 2}}{2 v^{- 2} z^{- 4}}$

Step 1

Move ${(3 z^{- 4})}^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 v^{- 2} z^{- 4} {(3 z^{- 4})}^{2}}$

Step 2

Move $v^{- 2}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$\frac{v^{2}}{2 z^{- 4} {(3 z^{- 4})}^{2}}$

Step 3

Move $z^{- 4}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$\frac{v^{2} z^{4}}{2 {(3 z^{- 4})}^{2}}$

Step 4

Simplify the denominator.

Apply the product rule to $3 z^{- 4}$ .

$\frac{v^{2} z^{4}}{2 \cdot 3^{2} {(z^{- 4})}^{2}}$

Raise $3$ to the power of $2$ .

$\frac{v^{2} z^{4}}{2 \cdot 9 {(z^{- 4})}^{2}}$

Multiply the exponents in ${(z^{- 4})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{v^{2} z^{4}}{2 \cdot 9 z^{- 4 \cdot 2}}$

Multiply $- 4$ by $2$ .

$\frac{v^{2} z^{4}}{2 \cdot 9 z^{- 8}}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{v^{2} z^{4}}{2 \cdot 9 \frac{1}{z^{8}}}$

Combine exponents.

Multiply $2$ by $9$ .

$\frac{v^{2} z^{4}}{18 \frac{1}{z^{8}}}$

Combine $18$ and $\frac{1}{z^{8}}$ .

$\frac{v^{2} z^{4}}{\frac{18}{z^{8}}}$

Step 5

Multiply the numerator by the reciprocal of the denominator.

$v^{2} z^{4} \frac{z^{8}}{18}$

Step 6

Multiply $v^{2} z^{4} \frac{z^{8}}{18}$ .

Combine $v^{2}$ and $\frac{z^{8}}{18}$ .

$z^{4} \frac{v^{2} z^{8}}{18}$

Combine $z^{4}$ and $\frac{v^{2} z^{8}}{18}$ .

$\frac{z^{4} (v^{2} z^{8})}{18}$

Multiply $z^{4}$ by $z^{8}$ by adding the exponents.

Move $z^{8}$ .

$\frac{z^{8} z^{4} v^{2}}{18}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{z^{8 + 4} v^{2}}{18}$

Add $8$ and $4$ .

$\frac{z^{12} v^{2}}{18}$