Question Number 193356 by leandrosriv02 last updated on 11/Jun/23
Answered by AST last updated on 11/Jun/23
$${m}=\mathrm{4} \\ $$$$\frac{{d}}{{dm}}\mathrm{3}^{{m}} =\mathrm{3}^{{m}} {In}\left(\mathrm{3}\right)>\:\frac{{d}}{{dm}}\left(\mathrm{2}^{{m}} +\mathrm{65}\right)=\mathrm{2}^{{m}} {In}\left(\mathrm{2}\right)\:\left({when}\:{m}>\mathrm{0}\right) \\ $$$${So},{there}\:{exists}\:{no}\:{other}\:{solution}\:{after}\:{their}\: \\ $$$${intersection}\:{at}\:{x}=\mathrm{4} \\ $$$${Suppose}\:{m}\leqslant\mathrm{0},{we}^{'} {d}\:{have}\:\mathrm{3}^{{m}} −\mathrm{2}^{{m}} \leqslant\mathrm{0}<\mathrm{65} \\ $$$$\Rightarrow{m}>\mathrm{0}\:{and}\:{only}\:{solution}\:{is}\:{m}=\mathrm{4} \\ $$
Answered by qaz last updated on 11/Jun/23
$${m}=\mathrm{2}{t} \\ $$$$\left(\mathrm{3}^{{t}} +\mathrm{2}^{{t}} \right)\left(\mathrm{3}^{{t}} −\mathrm{2}^{{t}} \right)=\mathrm{65}=\mathrm{5}×\mathrm{13} \\ $$$$\Rightarrow\begin{cases}{\mathrm{3}^{{t}} +\mathrm{2}^{{t}} =\mathrm{13}}\\{\mathrm{3}^{{t}} −\mathrm{2}^{{t}} =\mathrm{5}}\end{cases}\Rightarrow\mathrm{3}^{{t}} =\mathrm{9}\:\:\:,\mathrm{2}^{{t}} =\mathrm{4} \\ $$$$\Rightarrow{t}=\mathrm{2}\:\:,\:{m}=\mathrm{4} \\ $$