What is ((6*4^n)/(4^{2n+1)+2^{4n+1}})n^{-1} ?

The solution to ((6*4^n)/(4^{2n+1)+2^{4n+1}})n^{-1} is (3*2^{2n+1})/(n(4^{2n+1)+2^{4n+1})}

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simplify ((6*4^n)/(4^{2n+1)+2^{4n+1}})n^{-1}

simplify

(\frac{6 \cdot 4 ^{n}}{4 ^{2 n + 1} + 2 ^{4 n + 1}}) n^{- 1}

\frac{3 \cdot 2 ^{2 n + 1}}{n ( 4 ^{2 n + 1} + 2 ^{4 n + 1} )}

Solution steps

(\frac{6 \cdot 4 ^{n}}{4 ^{2 n + 1} + 2 ^{4 n + 1}}) n^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{n} \cdot \frac{6 \cdot 4 ^{n}}{4 ^{2 n + 1} + 2 ^{4 n + 1}}

\frac{6 \cdot 4 ^{n}}{4 ^{2 n + 1} + 2 ^{4 n + 1}} = \frac{3 \cdot 2 ^{1 + 2 n}}{4 ^{2 n + 1} + 2 ^{4 n + 1}}

= \frac{1}{n} \cdot \frac{3 \cdot 2 ^{2 n + 1}}{4 ^{2 n + 1} + 2 ^{4 n + 1}}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{3 \cdot 2 ^{1 + 2 n} \cdot 1}{( 4 ^{2 n + 1} + 2 ^{4 n + 1} ) n}

Multiply the numbers:

3 \cdot 1 = 3

= \frac{3 \cdot 2 ^{2 n + 1}}{n ( 4 ^{2 n + 1} + 2 ^{4 n + 1} )}

(x)^{\frac{5}{2}}

(a^{10} b^{0} c d e^{3})^{- 2}

- e^{- x}

0.4 e^{- 5000 (0.005)}

What is ((6*4^n)/(4^{2n+1)+2^{4n+1}})n^{-1} ?
The solution to ((6*4^n)/(4^{2n+1)+2^{4n+1}})n^{-1} is (3*2^{2n+1})/(n(4^{2n+1)+2^{4n+1})}