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Question Number 182575 by CrispyXYZ last updated on 11/Dec/22

$$\sqrt[x]{x+1}>\sqrt[e]{e}$$ $$1)\mathrm{Solve}\mathrm{for}x(x\in {\mathbb{N}}^{+}).$$ $$2)\mathrm{Solve}\mathrm{for}x(x>0).$$

Answered by mr W last updated on 12/Dec/22

$$f\left(x\right)={(1+x)}^{\frac{1}{x}}$$ $$forx>0:$$ $$f\left(x\right)isstrictlydecreasing.$$ $$f\left(x\right)={(1+x)}^{\frac{1}{x}}=e^{\frac{1}{e}}\Rightarrow x_1\approx 4.7591\left(solutionseelater\right)$$ $${(1+x)}^{\frac{1}{x}}>e^{\frac{1}{e}}$$ $$\Rightarrow 0<x<x_1\approx 4.7591$$ $$1)x\in [1,4]$$ $$2)x\in (0,4.7591)$$

Commented bymr W last updated on 12/Dec/22

$$howtosolve{(x+1)}^{\frac{1}{x}}=awitha>0$$ $$(x+1)=a^x$$ $$a(x+1)=a^{x+1}=e^{(x+1)\ln a}$$ $$-\ln a(x+1)e^{-(x+1)\ln a}=-\frac{\ln a}{a}$$ $$-\ln a(x+1)=\mathbb{W}(-\frac{\ln a}{a})$$ $$\Rightarrow x=-\frac{\mathbb{W}(-\frac{\ln a}{a})}{\ln a}-1$$ $$witha=e^{\frac{1}{e}},$$ $$x=-eW(-\frac{1}{e^{1+\frac{1}{e}}})-1$$ $$\approx -e\times (-2.118666)-1=4.7591or$$ $$\approx -e\times (-0.367879)-1=-1.1992\left(rejected\right)$$

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