Asymptotic time order

Question

Comments

$(A)?$

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Plz explain with solution .

Answer 1

$f(n)=n^{\frac{1}{\sqrt{\log n}}}= e^{\log n^{\frac{1}{\sqrt{\log n}}}}=e^{\sqrt{\log n}}$

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$g(n)=\sqrt{\log n}=e^{\log \sqrt{\log n}}$

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$h(n)=n^{\frac{1}{100}}=e^{\log n^{\frac{1}{100}} }=e^{k \times \log n}$

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$g(n)\, < \,f(n) \, <h(n)$

Comments

Thank u :)

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But I think if you compare f(n) and h(n), f(n) is greater because it will have 1/(logn)^1/2 term whereas h(n) will contain a constant term 1/100

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I am facing problems in logarithmic conversions. Like how you did the last step of simplifying f(n). Please suggest resources to read about these formulae.