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Question Number 170843 by cortano1 last updated on 01/Jun/22

Answered by floor(10²Eta[1]) last updated on 01/Jun/22

$$\frac{\log (\mathrm{x}-40)}{\log \sqrt{\mathrm{x}+1}+1}=1\iff \log (\mathrm{x}-40)=\log (\sqrt{\mathrm{x}+1}+1)$$$$\iff \mathrm{x}-40=\sqrt{\mathrm{x}+1}+1\iff \sqrt{\mathrm{x}+1}=\mathrm{x}-41$$$$\iff \mathrm{x}+1={(\mathrm{x}-41)}^2\wedge \mathrm{x}-41⩾0$$$$\iff {\mathrm{x}}^2-83\mathrm{x}+{41}^2-1=0\wedge \mathrm{x}⩾41$$$$\iff \mathrm{x}=\frac{83\pm \sqrt{169}}{2}\wedge \mathrm{x}⩾41$$$$\iff \mathrm{x}=\frac{83+13}{2}=48={\mathrm{x}}_0$$$$\Rightarrow \sqrt{{\mathrm{x}}_0+1}=\sqrt{49}=7$$

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