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