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Question Number 87550 by jagoll last updated on 05/Apr/20

$$\frac{1}{{2\mathrm{e}}^{-\mathrm{x}}-1}>\frac{2}{{\mathrm{e}}^{-\mathrm{x}}-2}$$

Commented byjohn santu last updated on 05/Apr/20

$$\Rightarrow \frac{{\mathrm{e}}^{\mathrm{x}}}{2-{\mathrm{e}}^{\mathrm{x}}}>\frac{{2\mathrm{e}}^{\mathrm{x}}}{1-{2\mathrm{e}}^{\mathrm{x}}}$$ $$[\mathrm{let}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}]$$ $$\frac{\mathrm{t}}{2-\mathrm{t}}>\frac{2\mathrm{t}}{1-2\mathrm{t}}\Rightarrow \frac{\mathrm{t}-{2\mathrm{t}}^2-4\mathrm{t}+{2\mathrm{t}}^2}{(2-\mathrm{t})(1-2\mathrm{t})}>0$$ $$\frac{-3\mathrm{t}}{(2-\mathrm{t})(1-2\mathrm{t})}>0$$ $$\Rightarrow \frac{1}{2}<\mathrm{t}<2$$ $${\mathrm{e}}^{\ln \left(\frac{1}{2}\right)}<{\mathrm{e}}^{\mathrm{x}}<{\mathrm{e}}^{\ln \left(2\right)}$$ $$-\ln \left(2\right)<\mathrm{x}<\ln \left(2\right)$$

Commented byjagoll last updated on 05/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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