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

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

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Simplify

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

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

Apply exponent rule:

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

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

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

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

4624

312 + 27 - 220

(- x - 6)^{2} (x^{2} - 12 x + 36)

\frac{\frac{2}{7}}{\frac{3}{5} - \frac{3}{10}}

\frac{cos ( x )}{- sin ( x )}

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