Menu Close

Find-the-value-s-of-a-such-that-a-x-ax-with-a-x-R-




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

Leave a Reply

Your email address will not be published. Required fields are marked *