Question Number 45982 by MrW3 last updated on 19/Oct/18
$${Find}\:{the}\:{value}\left({s}\right)\:{of}\:{a}\:{such}\:{that} \\ $$$${a}^{{x}} \geqslant{ax}\:{with}\:{a},\:{x}\in{R}. \\ $$
Commented by ajfour last updated on 19/Oct/18
Answered by ajfour last updated on 19/Oct/18
$$\underset{−−−−−−−} {{For}\:{x}\:\geqslant\:\mathrm{1}} \\ $$$${x}\geqslant\mathrm{1}+\mathrm{ln}\:_{{a}} {x} \\ $$$$\mathrm{ln}\:_{{a}} {x}\:\leqslant\:{x}−\mathrm{1} \\ $$$$\mathrm{ln}\:{x}\:\leqslant\:\left(\mathrm{ln}\:{a}\right)\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{1}}{{x}}\mid_{{x}=\mathrm{1}} \:\leqslant\:\:\mathrm{ln}\:{a}\:\:\:\Rightarrow\:{a}\:\geqslant\:{e} \\ $$$$\underset{−−−−−−} {{For}\:{x}\:\leqslant\:\mathrm{1}} \\ $$$$\mathrm{ln}\:{x}\:\leqslant\:\left(\mathrm{ln}\:{a}\right)\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\frac{{d}}{{dx}}\left(\mathrm{ln}\:{x}\right)\mid_{{x}=\mathrm{1}} \:\geqslant\:\mathrm{ln}\:{a} \\ $$$$\Rightarrow\:\:\mathrm{ln}\:{a}\:\leqslant\:\mathrm{1}\:\:\:\Rightarrow\:\:\:\boldsymbol{{a}}\:\leqslant\:\boldsymbol{{e}} \\ $$$$\boldsymbol{{Hence}}\:\boldsymbol{{for}}\:\boldsymbol{{all}}\:\boldsymbol{{x}}\:\in{R}\:,\:\boldsymbol{{a}}\:>\mathrm{0} \\ $$$$\:\:\:\boldsymbol{{a}}^{\boldsymbol{{x}}} \geqslant\:\boldsymbol{{ax}}\:\:\:{if}\:\:\:\boldsymbol{{a}}=\boldsymbol{{e}}\:. \\ $$
Commented by ajfour last updated on 19/Oct/18
$${please}\:{check}\:{again}\:{Sir},\:{i}\:{have} \\ $$$${edited}\:{though}\:{i}\:{have}\:{not}\: \\ $$$${explained}.. \\ $$
Commented by MrW3 last updated on 19/Oct/18
$${that}'{s}\:{perfect}\:{sir}!\:{thanks}\:{alot}! \\ $$