Simplify (4/(4^-3))^-2

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

Step 1

Move $4^{- 3}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

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

Step 2

Multiply $4$ by $4^{3}$ by adding the exponents.

Multiply $4$ by $4^{3}$ .

Raise $4$ to the power of $1$ .

${(4^{1} \cdot 4^{3})}^{- 2}$

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

${(4^{1 + 3})}^{- 2}$

Add $1$ and $3$ .

${(4^{4})}^{- 2}$

Step 3

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

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

$4^{4 \cdot - 2}$

Multiply $4$ by $- 2$ .

$4^{- 8}$

Step 4

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

$\frac{1}{4^{8}}$

Step 5

Raise $4$ to the power of $8$ .

$\frac{1}{65536}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{65536}$

Decimal Form:

$1.5258789 \cdot 10^{- 5}$