What is n^{1/(2^{log_{2)(log_{2}(n))}}} ?

The solution to n^{1/(2^{log_{2)(log_{2}(n))}}} is n^{1/(log_{2)(n)}}

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simplify n^{1/(2^{log_{2)(log_{2}(n))}}}

simplify

n^{\frac{1}{2 ^{l o g_{2} (l o g_{2} (n))}}}

n^{\frac{1}{l o g _{2} ( n )}}

Solution steps

n^{\frac{1}{2 ^{l o g_{2} (l o g_{2} (n))}}}

Apply log rule:

a^{l o g_{a} (b)} = b

2^{l o g_{2} (l o g_{2} (n))} = lo g_{2} (n)

= n^{\frac{1}{l o g _{2} ( n )}}

20 (\frac{π}{4})

\frac{3}{4} (x - 1) - \frac{1}{2} (3 x + 2) = 0

\frac{2 ^{n + 1}}{2 ^{n}}

1.2 = \frac{w}{- 6}

\frac{29}{5 + 2 i}

What is n^{1/(2^{log_{2)(log_{2}(n))}}} ?
The solution to n^{1/(2^{log_{2)(log_{2}(n))}}} is n^{1/(log_{2)(n)}}