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  1. Prove or disprove that $\log_{10}{\sqrt{n}} = \Theta(\lg{n^{5}})$
  2. Prove or disprove that $4^{\log_{10}{n}}=O(\sqrt{n}\lg{n^{3}})$.

For the first question, I divide $\lg{n^{5}}$ on both side and find that $$\frac{\log_{10}{\sqrt{n}}}{\lg{n^{5}}} = \frac{\frac{1}{2}\log_{10}{n}}{5\log_{2}{n}} = \frac{1}{10}\frac{\log_{n}{2}}{\log_{n}{10}} = \frac{1}{10}\log_{10}{2},$$ which is approximately equal to 0.03. Hence I can find $c_{1} = 0.02,c_{2} = 0.1$ such that $c_{1}\lg{n^{5}} \leq \log_{10}{\sqrt{n}} \leq c_{2}\lg{n^{5}}, \forall n \geq 1$. And I don’t know whether this proof is right?

For the second question, I tried to follow the similar logic as the first question, but after dividing $\sqrt{n}\lg{n^{3}}$ on both sides, I got stuck. Trying to compute the derivative of $\frac{4^{\log_{10}{n}}}{\sqrt{n}\lg{n^{3}}}$ and find the asymptotic properties is just too time-consuming and doesn’t necessarily lead to the proof or disproof, so can anyone help me on this?

Rócherz
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  • You can use \log and \lg to produce $\log$ and $\lg$ instead of the not so great looking $log$ and $lg$. I corrected them for you this time. – Gary Oct 24 '21 at 03:29
  • First solution is correct. For the second one see my comments for the same problem from yesterday: https://math.stackexchange.com/q/4283930 – Gary Oct 24 '21 at 03:32

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