Question Number 647 by 123456 last updated on 18/Feb/15 $$\mathrm{log}_{{x}} \left({y}^{\pi} \right)+\mathrm{log}_{{y}} \left({x}^{{e}} \right)={a} \\ $$$$\frac{\mathrm{1}}{\mathrm{log}_{{y}} \left({x}^{\pi^{−\mathrm{1}} } \right)}−\frac{\mathrm{1}}{\mathrm{log}_{{x}} \left({y}^{{e}^{−\mathrm{1}} } \right)}={b} \\ $$$$\frac{{x}^{{a}+{b}+\mathrm{2}{e}} }{{y}^{{a}−{b}+\mathrm{2}\pi}…
Question Number 131697 by mathlove last updated on 07/Feb/21 $${x}^{{x}} =\mathrm{6}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$ Answered by mr W last updated on 07/Feb/21 $${x}=\mathrm{6}^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{ln}\:\mathrm{6}}{{x}}} \\ $$$$\frac{\mathrm{1}}{{x}}{e}^{\frac{\mathrm{ln}\:\mathrm{6}}{{x}}}…
Question Number 619 by Cheenz last updated on 11/Feb/15 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:+\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:=\:? \\ $$ Answered by 123456 last updated on 11/Feb/15 $${x}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}+\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{2}}…
Question Number 131688 by liberty last updated on 07/Feb/21 $$\:\mathrm{Let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{2}=\mathrm{0}. \\ $$$$\mathrm{If}\:{a}_{{n}} \:=\:\alpha^{{n}} −\beta^{{n}} \:\mathrm{for}\:{n}\:\geqslant\mathrm{1}\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{{a}_{\mathrm{10}} −\mathrm{2}{a}_{\mathrm{8}} }{\mathrm{2}{a}_{\mathrm{9}} }\:? \\ $$…
Question Number 66126 by Joel122 last updated on 09/Aug/19 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{find}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{1}\:+\:{n}^{\mathrm{2}} \:+\:{n}^{\mathrm{4}} \:+\:{n}^{\mathrm{6}} \:+\:{n}^{\mathrm{8}} \:+\:…\:+\:{n}^{\mathrm{2}{k}} \:+\:… \\ $$$$\mathrm{where}\:{n},{k}\:\in\:\mathbb{Z}^{+} \: \\ $$ Answered by mr…
Question Number 588 by shaleen last updated on 03/Feb/15 $$\left({a}+{b}=\mathrm{10}\right)\:\left({a}+{c}=\mathrm{24}\right)\:\left({b}+{c}=\mathrm{20}\right) \\ $$$${find}\:{a}\:{b}\:{c} \\ $$ Answered by prakash jain last updated on 03/Feb/15 $${a}+{b}=\mathrm{10}\Rightarrow{b}=\mathrm{10}−{a} \\ $$$${a}+{c}=\mathrm{24}\Rightarrow{c}=\mathrm{24}−{a}…
Question Number 577 by Bek last updated on 31/Jan/15 $${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$ Answered by Bek last updated on 31/Jan/15 $$ \\ $$…
Question Number 131640 by Salman_Abir last updated on 07/Feb/21 Answered by EDWIN88 last updated on 07/Feb/21 $$ \\ $$$$\mathrm{Du}\:\mathrm{bist}\:\mathrm{wundersch}\ddot {\mathrm{o}n} \\ $$ Commented by EDWIN88…
Question Number 131605 by physicstutes last updated on 06/Feb/21 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\:{m}^{\mathrm{4}} −\mathrm{7}{m}^{\mathrm{3}} +\:\mathrm{14}{m}^{\mathrm{2}} −\mathrm{7}{m}\:+\:\mathrm{1}\:=\:\mathrm{0}\: \\ $$ Answered by malwan last updated on 06/Feb/21 $${m}^{\mathrm{4}}…
Question Number 482 by prakash jain last updated on 12/Jan/15 $$\mathrm{The}\:\mathrm{number}\:\mathrm{1000}!\:\mathrm{has}\:\mathrm{certain}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{0}{s}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end},\:\mathrm{what}\:\mathrm{the}\:\mathrm{the}\:\mathrm{first}\:\mathrm{non}−\mathrm{zero} \\ $$$$\mathrm{digit}. \\ $$$$\mathrm{1000}!=…\mathrm{d}_{\mathrm{1}} \mathrm{d}_{\mathrm{2}} \mathrm{d}_{\mathrm{3}} \mathrm{D00000}… \\ $$$$\mathrm{where}\:\mathrm{d}_{\mathrm{1}} ,\mathrm{d}_{\mathrm{2}} ,\mathrm{d}_{\mathrm{3}} ,\mathrm{D}\:\mathrm{are}\:\mathrm{digits}.…