Question Number 191392 by Mingma last updated on 23/Apr/23
Answered by Frix last updated on 23/Apr/23
$$\frac{\sqrt{\mathrm{ln}\:{n}}}{\:\sqrt{\mathrm{ln}\:{b}}}=\frac{\mathrm{ln}\:{n}}{\mathrm{2ln}\:{b}}\:\Rightarrow\:\mathrm{ln}\:{n}\:=\mathrm{4ln}\:{b} \\ $$$$\frac{{b}\mathrm{ln}\:{n}}{\mathrm{ln}\:{b}}=\frac{\mathrm{ln}\:{bn}}{\mathrm{ln}\:{b}}\:\Rightarrow\:\mathrm{ln}\:{n}\:=\frac{\mathrm{ln}\:{b}}{{b}−\mathrm{1}} \\ $$$$\Rightarrow \\ $$$$\mathrm{4ln}\:{b}\:=\frac{\mathrm{ln}\:{b}}{{b}−\mathrm{1}}\:\Rightarrow\:{b}=\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow\:{n}=\frac{\mathrm{625}}{\mathrm{256}} \\ $$
Commented by Mingma last updated on 24/Apr/23
Excellent
Answered by cortano12 last updated on 24/Apr/23
$$\:\mathrm{let}\:\mathrm{log}\:_{\mathrm{b}} \mathrm{n}\:=\:\mathrm{x} \\ $$$$\:\begin{cases}{\sqrt{\mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\Rightarrow\mathrm{4x}=\mathrm{x}^{\mathrm{2}} \:\Rightarrow\begin{cases}{\mathrm{x}=\mathrm{0}}\\{\mathrm{x}=\mathrm{4}}\end{cases}}\\{\mathrm{bx}=\mathrm{1}+\mathrm{x}\Rightarrow\mathrm{4b}=\mathrm{5}\:;\:\mathrm{b}=\frac{\mathrm{5}}{\mathrm{4}}}\end{cases} \\ $$$$\:\Leftrightarrow\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{4}}} \mathrm{n}\:=\:\mathrm{4}\:;\:\mathrm{n}=\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{4}} =\frac{\mathrm{625}}{\mathrm{256}} \\ $$
Commented by Mingma last updated on 24/Apr/23
Excellent