Simplify ((2p^5m)^-2)/p

$\frac{{(2 p^{5} m)}^{- 2}}{p}$

Step 1

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

$\frac{1}{p {(2 p^{5} m)}^{2}}$

Step 2

Simplify the denominator.

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

$\frac{1}{p {(2 p^{5})}^{2} m^{2}}$

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

$\frac{1}{p (2^{2} {(p^{5})}^{2}) m^{2}}$

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

$\frac{1}{p (4 {(p^{5})}^{2}) m^{2}}$

Multiply the exponents in ${(p^{5})}^{2}$ .

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

$\frac{1}{p (4 p^{5 \cdot 2}) m^{2}}$

Multiply $5$ by $2$ .

$\frac{1}{p (4 p^{10}) m^{2}}$

$\frac{1}{p \cdot 4 p^{10} m^{2}}$

Multiply $p$ by $p^{10}$ by adding the exponents.

Move $p^{10}$ .

$\frac{1}{p^{10} p \cdot 4 m^{2}}$

Multiply $p^{10}$ by $p$ .

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

$\frac{1}{p^{10} p^{1} \cdot 4 m^{2}}$

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

$\frac{1}{p^{10 + 1} \cdot 4 m^{2}}$

Add $10$ and $1$ .

$\frac{1}{p^{11} \cdot 4 m^{2}}$

Step 3

Move $4$ to the left of $p^{11}$ .

$\frac{1}{4 p^{11} m^{2}}$