Question Number 222323 by MrGaster last updated on 22/Jun/25
$${a}+\mathrm{3}^{{b}} ={b},\mathrm{3}^{{b}} \centerdot{a}^{{b}+\mathrm{1}} \:\mathrm{max}=? \\ $$
Answered by Frix last updated on 22/Jun/25
$${a}={b}−\mathrm{3}^{{b}} \\ $$$${f}\left({b}\right)=\mathrm{3}^{{b}} \left({b}−\mathrm{3}^{{b}} \right)^{{b}+\mathrm{1}} \\ $$$${f}\left({b}−\mathrm{1}\right)=\frac{\left(\mathrm{3}{b}−\mathrm{3}−\mathrm{3}^{{b}} \right)^{{b}} }{\mathrm{3}} \\ $$$$\mathrm{3}{b}−\mathrm{3}−\mathrm{3}^{{b}} <\mathrm{0}\forall{b}\in\mathbb{R}\:\Rightarrow\:{f}\left({b}−\mathrm{1}\right)\in\mathbb{R}\:\Leftrightarrow\:{b}\in\mathbb{Z} \\ $$$${b}=\mathrm{2}{n}:\:{n}\rightarrow+\infty\:\Rightarrow\:{f}\left({b}−\mathrm{1}\right)\rightarrow+\infty \\ $$$${b}=\mathrm{2}{n}+\mathrm{1}:\:{n}\rightarrow+\infty\:\Rightarrow\:{f}\left({b}−\mathrm{1}\right)\rightarrow−\infty \\ $$$${n}\rightarrow−\infty\:\Rightarrow\:{f}\left({b}−\mathrm{1}\right)\rightarrow\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{no}\:\mathrm{maximum}\:\mathrm{exists} \\ $$