Solve for L L=(2^(n+1)*4^(-2n+1)+8^(-n+2))/(16(2^n)^-3)

$L = \frac{2^{n + 1} \cdot 4^{- 2 n + 1} + 8^{- n + 2}}{16 {(2^{n})}^{- 3}}$

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

$L = \frac{(2^{n + 1} \cdot 4^{- 2 n + 1} + 8^{- n + 2}) {(2^{n})}^{3}}{16}$

Simplify the numerator.

Multiply $2^{n + 1} \cdot 4^{- 2 n + 1}$ .

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

$L = \frac{(2^{n + 1} \cdot {(2^{2})}^{- 2 n + 1} + 8^{- n + 2}) {(2^{n})}^{3}}{16}$

Multiply the exponents in ${(2^{2})}^{- 2 n + 1}$ .

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

$L = \frac{(2^{n + 1} \cdot 2^{2 (- 2 n + 1)} + 8^{- n + 2}) {(2^{n})}^{3}}{16}$

Apply the distributive property.

$L = \frac{(2^{n + 1} \cdot 2^{2 (- 2 n) + 2 \cdot 1} + 8^{- n + 2}) {(2^{n})}^{3}}{16}$

Multiply $- 2$ by $2$ .

$L = \frac{(2^{n + 1} \cdot 2^{- 4 n + 2 \cdot 1} + 8^{- n + 2}) {(2^{n})}^{3}}{16}$

Multiply $2$ by $1$ .

$L = \frac{(2^{n + 1} \cdot 2^{- 4 n + 2} + 8^{- n + 2}) {(2^{n})}^{3}}{16}$

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

$L = \frac{(2^{n + 1 - 4 n + 2} + 8^{- n + 2}) {(2^{n})}^{3}}{16}$

Subtract $4 n$ from $n$ .

$L = \frac{(2^{- 3 n + 1 + 2} + 8^{- n + 2}) {(2^{n})}^{3}}{16}$

Add $1$ and $2$ .

$L = \frac{(2^{- 3 n + 3} + 8^{- n + 2}) {(2^{n})}^{3}}{16}$

Multiply the exponents in ${(2^{n})}^{3}$ .

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

$L = \frac{(2^{- 3 n + 3} + 8^{- n + 2}) \cdot 2^{n \cdot 3}}{16}$

Move $3$ to the left of $n$ .

$L = \frac{(2^{- 3 n + 3} + 8^{- n + 2}) \cdot 2^{3 n}}{16}$

Reorder factors in $\frac{(2^{- 3 n + 3} + 8^{- n + 2}) \cdot 2^{3 n}}{16}$ .

$L = \frac{2^{3 n} (2^{- 3 n + 3} + 8^{- n + 2})}{16}$