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Question Number 191501 by normans last updated on 24/Apr/23
      find a solution;              e^x  = ln(x)
$$ \\ $$$$\:\:\:\:{find}\:{a}\:{solution}; \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{e}}^{\boldsymbol{{x}}} \:=\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right) \\ $$$$ \\ $$
Commented by mehdee42 last updated on 24/Apr/23
this equation has no solution
$${this}\:{equation}\:{has}\:{no}\:{solution} \\ $$
Commented by mehdee42 last updated on 24/Apr/23
Answered by Frix last updated on 25/Apr/23
e^x =ln x ⇔ x=e^x =ln x  x=e^x   (−x)e^(−x) =−1  x≈.318131505205±1.33723570143i
$$\mathrm{e}^{{x}} =\mathrm{ln}\:{x}\:\Leftrightarrow\:{x}=\mathrm{e}^{{x}} =\mathrm{ln}\:{x} \\ $$$${x}=\mathrm{e}^{{x}} \\ $$$$\left(−{x}\right)\mathrm{e}^{−{x}} =−\mathrm{1} \\ $$$${x}\approx.\mathrm{318131505205}\pm\mathrm{1}.\mathrm{33723570143i} \\ $$
Commented by mehdee42 last updated on 28/Apr/23
?! e^x =lnx⇔x=e^x =lnx ?!
$$?!\:{e}^{{x}} ={lnx}\Leftrightarrow{x}={e}^{{x}} ={lnx}\:?! \\ $$
Commented by Frix last updated on 13/May/23
f(x)=f^(−1) (x) ⇒ x=f(x)=f^(−1) (x)
$${f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\:\Rightarrow\:{x}={f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right) \\ $$

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