Simplify ((2k^5)^-3)/(2j^5k^3)

$\frac{{(2 k^{5})}^{- 3}}{2 j^{5} k^{3}}$

Step 1

Move ${(2 k^{5})}^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 j^{5} k^{3} {(2 k^{5})}^{3}}$

Step 2

Simplify the denominator.

Apply the product rule to $2 k^{5}$ .

$\frac{1}{2 j^{5} k^{3} \cdot 2^{3} {(k^{5})}^{3}}$

Multiply $2$ by $2^{3}$ by adding the exponents.

Move $2^{3}$ .

$\frac{1}{2^{3} \cdot 2 j^{5} k^{3} {(k^{5})}^{3}}$

Multiply $2^{3}$ by $2$ .

Raise $2$ to the power of $1$ .

$\frac{1}{2^{3} \cdot 2^{1} j^{5} k^{3} {(k^{5})}^{3}}$

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

$\frac{1}{2^{3 + 1} j^{5} k^{3} {(k^{5})}^{3}}$

Add $3$ and $1$ .

$\frac{1}{2^{4} j^{5} k^{3} {(k^{5})}^{3}}$

Raise $2$ to the power of $4$ .

$\frac{1}{16 j^{5} k^{3} {(k^{5})}^{3}}$

Multiply the exponents in ${(k^{5})}^{3}$ .

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

$\frac{1}{16 j^{5} k^{3} k^{5 \cdot 3}}$

Multiply $5$ by $3$ .

$\frac{1}{16 j^{5} k^{3} k^{15}}$

Multiply $k^{3}$ by $k^{15}$ by adding the exponents.

Move $k^{15}$ .

$\frac{1}{16 j^{5} (k^{15} k^{3})}$

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

$\frac{1}{16 j^{5} k^{15 + 3}}$

Add $15$ and $3$ .

$\frac{1}{16 j^{5} k^{18}}$