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Question Number 156107 by cortano last updated on 08/Oct/21

$${(1+\frac{1}{\mathrm{x}})}^{\mathrm{x}+1}={(1+\frac{1}{2019})}^{2019}$$

Commented by mr W last updated on 08/Oct/21

$${(1+\frac{1}{x})}^{x+1}$$$$={\left(\frac{x+1}{x}\right)}^{x+1}$$$$={\left(\frac{x}{x+1}\right)}^{-(x+1)}$$$$={\left(\frac{x+1-1}{x+1}\right)}^{-(x+1)}$$$$={(1-\frac{1}{x+1})}^{-(x+1)}$$$$={(1+\frac{1}{-(x+1)})}^{-(x+1)}={(1+\frac{1}{n})}^n$$$$\Rightarrow -(x+1)=n$$$$\Rightarrow x=-(1+n)$$

Answered by VIDDD last updated on 08/Oct/21

$$Ans...$$$$find\mathbf{x}$$$${(1+\frac{1}{\mathbf{x}})}^{\mathbf{x}+1}={(1+\frac{1}{2019})}^{2019}$$$$\mathrm{let}\mathrm{y}=\mathrm{x}+1;\mathrm{x}=\mathrm{y}-1\left(\mathrm{\$}\right)$$$$\Rightarrow {(1+\frac{1}{\mathrm{y}-1})}^{\mathrm{y}}={(1+\frac{1}{2019})}^{2019}$$$$\mathrm{change}\mathrm{y}=-\mathrm{z}.$$$$\Rightarrow {(1+\frac{1}{-\mathrm{z}-1})}^{-\mathrm{z}}={(1+\frac{1}{2019})}^{2019}$$$${\left(\frac{\mathrm{z}}{\mathrm{z}+1}\right)}^{-\mathrm{z}}={(1+\frac{1}{2019})}^{2019}$$$${\left(\frac{\mathrm{z}+1}{\mathrm{z}}\right)}^{\mathrm{z}}={(1+\frac{1}{2019})}^{2019}$$$${(1+\frac{1}{\mathrm{z}})}^{\mathrm{z}}={(1+\frac{1}{2019})}^{2019}$$$$\iff \mathrm{z}=2019;\mathrm{but}\mathrm{y}=-\mathrm{z}\Rightarrow \mathrm{y}=-2019$$$$\left(\mathrm{\$}\right)\mathbf{x}=-2019-1=-2020$$$$\mathit{S}\mathit{o}\mathbf{x}=-2020\mathbf{vid}$$$$$$

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