Basic Math Examples

$\frac{8^{- n + 1} - 4^{- n} \cdot 2^{n}}{4 {(2^{2 n})}^{- 1}}$

Step 1

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

$\frac{(8^{- n + 1} - 4^{- n} \cdot 2^{n}) \cdot 2^{2 n}}{4}$

Step 2

Multiply $- 4^{- n} \cdot 2^{n}$ .

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Step 2.1

Rewrite $4$ as $2^{2}$ .

$\frac{(8^{- n + 1} - (2^{n} \cdot {(2^{2})}^{- n})) \cdot 2^{2 n}}{4}$

Step 2.2

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

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Step 2.2.1

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

$\frac{(8^{- n + 1} - (2^{n} \cdot 2^{2 (- n)})) \cdot 2^{2 n}}{4}$

Step 2.2.2

Multiply $- 1$ by $2$ .

$\frac{(8^{- n + 1} - (2^{n} \cdot 2^{- 2 n})) \cdot 2^{2 n}}{4}$

Step 2.3

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

$\frac{(8^{- n + 1} - 2^{n - 2 n}) \cdot 2^{2 n}}{4}$

Step 2.4

Subtract $2 n$ from $n$ .

$\frac{(8^{- n + 1} - 2^{- n}) \cdot 2^{2 n}}{4}$

Step 3

Reorder factors in $\frac{(8^{- n + 1} - 2^{- n}) \cdot 2^{2 n}}{4}$ .

$\frac{2^{2 n} (8^{- n + 1} - 2^{- n})}{4}$