Prove by PMI
logx^n=nlogx

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{P}\left(\mathrm{n}\right):{\mathrm{logx}}^{\mathrm{n}}=\mathrm{nlogx} \\ \mathrm{Step} 1: \mathrm{Base} \mathrm{Case} \\ \mathrm{For} \mathrm{n}=1 \\ {\mathrm{logx}}^1=\mathrm{logx} \\ \Rightarrow \mathrm{logx}=\mathrm{logx} \\ \mathrm{which} \mathrm{is} \mathrm{true}. \\ \mathrm{Hence} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=1. \\ \mathrm{Step} 2: \mathrm{Inductive} \mathrm{hypothesis} \\ \mathrm{Let} \mathrm{P}\left(\mathrm{n}\right) \mathrm{be} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k} \\ \Rightarrow {\mathrm{logx}}^{\mathrm{k}}=\mathrm{klogx} ;\left(\mathrm{i}\right) \\ \mathrm{Step} 3: \mathrm{Inductive} \mathrm{Case} \\ \mathrm{Consider} \\ \log \left({\mathrm{x}}^{\mathrm{k}+1}\right)=\log \left({\mathrm{x}}^{\mathrm{k}}.\mathrm{x}\right) \\ ={\mathrm{logx}}^{\mathrm{k}}+\mathrm{logx} \left[\mathrm{As} \log \left(\mathrm{ab}\right)=\log \left(\mathrm{a}\right)+\log \left(\mathrm{b}\right)\right] \\ =\mathrm{klogx}+\mathrm{logx} \left[\mathrm{Using} \left(\mathrm{i}\right)\right] \\ =\left(\mathrm{k}+1\right)\mathrm{logx} \\ \Rightarrow \log \left({\mathrm{x}}^{\mathrm{k}+1}\right)=\left(\mathrm{k}+1\right)\mathrm{logx} \\ \mathrm{Hence} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k}+1 \\ \mathrm{Since} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=1 \mathrm{and} \mathrm{assumption} \mathrm{of} \mathrm{truth} \mathrm{for} \mathrm{n}=\mathrm{k}, \mathrm{gives} \mathrm{us} \mathrm{the} \mathrm{result} \\ \mathrm{that} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{also} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k}+1. \\ \mathrm{Hence} \mathrm{by} \mathrm{Principal} \mathrm{of} \mathrm{mathematical} \mathrm{induction} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true}. \end{gathered}$$

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