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Question Number 115696 by Dwaipayan Shikari last updated on 27/Sep/20

$${\mathrm{e}}^{\mathrm{x}}=\mathrm{logx}$$

Commented by Dwaipayan Shikari last updated on 27/Sep/20

$${\mathrm{e}}^{\mathrm{x}}=\mathrm{logx}$$$${\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}=\mathrm{x}$$$${\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{d}}^{{\mathrm{d}}^{\mathrm{d}}}}}}}=\mathrm{x}$$$${\mathrm{e}}^{\mathrm{x}}=\mathrm{x}$$$$\mathrm{x}=\mathrm{logx}$$$${\mathrm{e}}^{-\mathrm{logx}}=\frac{1}{\mathrm{logx}}$$$$-{\mathrm{logxe}}^{-\mathrm{logx}}=-$$$$-\mathrm{logx}={\mathrm{W}}_0(-1)$$$$\mathrm{x}={\mathrm{e}}^{-{\mathrm{W}}_0(-1)}(\mathrm{Complex}\mathrm{solution}\mathrm{x}\in \mathbb{C})$$$$\mathrm{Error}!$$$${\mathrm{e}}^{\mathrm{x}}=\mathrm{logx}$$$$\mathrm{And}{\mathrm{e}}^{\mathrm{x}}=\mathrm{x}\left({\mathrm{dosen}}^{\prime }\mathrm{t}\mathrm{satisfy}\mathrm{any}\mathrm{condition}\right)$$$$$$

Commented by TANMAY PANACEA last updated on 27/Sep/20

$$fromgraphitisclearthat{e}^{x}andlnxnevercross$$$$eachother...sonosolution$$

Commented by Dwaipayan Shikari last updated on 27/Sep/20

$$\mathrm{Complex}\mathrm{solution}$$$$\mathrm{No}\mathrm{real}\mathrm{solution}$$

Commented by TANMAY PANACEA last updated on 27/Sep/20

$$ok$$

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