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Question Number 95485 by O Predador last updated on 25/May/20

Commented by PRITHWISH SEN 2 last updated on 25/May/20

$${2\mathrm{llog}}_{\ln \left(\pi \right)}\frac{1}{\sqrt{\mathrm{x}}+\sqrt{\ln \left(\pi \right)}}-{2\log }_{\ln \left(\pi \right)}\frac{1}{\mathrm{x}-\mathrm{lnx}}=1$$$${2\log }_{\ln \left(\pi \right)}\frac{\mathrm{x}-\ln \pi }{\sqrt{\mathrm{x}}+\sqrt{\ln \left(\pi \right)}}=1$$$${\log }_{\ln \pi }{(\sqrt{\mathrm{x}}+\sqrt{\ln \pi })}^2=1$$$$\mathrm{x}+2\sqrt{\mathrm{x}\left(\ln \pi \right)}+\ln \pi =\ln \pi$$$$\sqrt{\mathrm{x}}=-2\sqrt{\ln \pi }\{\because \mathrm{x}\ne 0\}$$$$\mathbf{x}=4\mathbf{ln}\mathit{\pi }$$$$$$

Commented by O Predador last updated on 25/May/20

$$\mathrm{Mito}!$$

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