What is the answer to n^{1.01}=o(nlog_{2}(n)) ?

The answer to n^{1.01}=o(nlog_{2}(n)) is o=(n^{0.01})/(log_{2)(n)};\quad n\ne 1

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n^{1.01}=o(nlog_{2}(n))

n^{1.01} = o (n lo g_{2} (n))

o = \frac{n ^{0.01}}{lo g _{2} ( n )}; n \neq = 1

Solution steps

n^{1.01} = o (n lo g_{2} (n))

Switch sides

o (n lo g_{2} (n)) = n^{1.01}

Divide both sides by

n lo g_{2} (n); n \neq = 1

o = \frac{n ^{0.01}}{lo g _{2} ( n )}; n \neq = 1

8.9 + 1.2 (3 a - 1) = 14.9

x, x y + y^{3} = 2

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

\frac{x}{2} + 13 = 20

a, 15 a + 5 b = 60

What is the answer to n^{1.01}=o(nlog_{2}(n)) ?
The answer to n^{1.01}=o(nlog_{2}(n)) is o=(n^{0.01})/(log_{2)(n)};\quad n\ne 1