Question Number 92135 by jagoll last updated on 05/May/20
$${a}^{{x}} \:=\:\mathrm{log}\:_{{a}} \:\left({x}\right) \\ $$$${a}=? \\ $$
Commented by Tony Lin last updated on 05/May/20
$${when}\:\mathrm{0}<{x}<{e}^{−{e}} ,\:{x}\:{has}\:{three}\:{solutions} \\ $$$${when}\:{e}^{−{e}} \leqslant{a}<\mathrm{1},\:{x}\:{has}\:{one}\:{solution} \\ $$$${when}\:\mathrm{1}<{a}<\:{e}^{\frac{\mathrm{1}}{{e}}} ,\:{x}\:{has}\:{two}\:{solutions} \\ $$$${when}\:{a}={e}^{\frac{\mathrm{1}}{{e}}} ,\:{x}\:{has}\:{one}\:{solution} \\ $$$${when}\:{a}>{e}^{\frac{\mathrm{1}}{{e}}} ,\:{x}\:{has}\:{no}\:{solutions} \\ $$$$\therefore\:{when}\:{a}^{{x}} ={log}_{{a}} {x} \\ $$$$\Rightarrow{a}\epsilon\left(\mathrm{0},{e}^{\frac{\mathrm{1}}{{e}}} \right]\backslash\left\{\mathrm{1}\right\} \\ $$
Commented by mr W last updated on 05/May/20
$${a}=\mathrm{1}\:\Rightarrow{one}\:{solution}\:{x}=\mathrm{1} \\ $$
Commented by jagoll last updated on 05/May/20
$$\mathrm{log}\:_{\mathrm{1}} \left(\mathrm{1}\right)\:\mathrm{defined}? \\ $$
Commented by MJS last updated on 05/May/20
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\mathrm{log}_{{x}} \:{x}\:=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}}\:=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{1}\:=\mathrm{1} \\ $$$$\mathrm{also}\:\mathrm{log}_{\mathrm{0}} \:\mathrm{0}\:=\mathrm{1} \\ $$$$\mathrm{log}_{{z}} \:{z}\:=\mathrm{1}\:\forall{z}\in\mathbb{C} \\ $$
Commented by Tony Lin last updated on 05/May/20
$${a}^{{a}^{{x}} } ={x} \\ $$$${a}^{{x}} {lna}={lnx} \\ $$$$\left({xlna}\right){e}^{{xlna}} ={xlnx} \\ $$$${xlna}=\mathbb{W}\left({xlnx}\right) \\ $$$${a}={e}^{\frac{\mathbb{W}\left({xlnx}\right)}{{x}}} \\ $$