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Author: Tinku Tara

Question-217733

Question Number 217733 by Samuel12 last updated on 19/Mar/25 Answered by vnm last updated on 19/Mar/25 $${x}^{\frac{\mathrm{1}}{\mathrm{ln}\:\left({e}^{{x}} −\mathrm{1}\right)}} ={e}^{\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\left({e}^{{x}} −\mathrm{1}\right)}} ={e}^{\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}+\mathrm{ln}\:\frac{{e}^{{x}} −\mathrm{1}}{{x}}}} = \\ $$$${e}^{\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{ln}\:{x}}\mathrm{ln}\:\frac{{e}^{{x}}…