Question Number 15570 by tawa tawa last updated on 11/Jun/17 $$\mathrm{Solve}:\:\:\:\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{x}^{\mathrm{4}} \\ $$ Commented by tawa tawa last updated on 11/Jun/17 $$\mathrm{workings} \\ $$…
Question Number 146636 by mathdanisur last updated on 14/Jul/21 $$\mathrm{4}\:+\:\frac{\mathrm{12}}{\mathrm{8}\:+\:\frac{\mathrm{7}}{\mathrm{3}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}\:+\:\mathrm{1}}}}\:=\:\mathrm{8}\:\:\Rightarrow\:\:\boldsymbol{{x}}=? \\ $$ Commented by hknkrc46 last updated on 14/Jul/21 $$\bigstar\:\frac{\mathrm{12}}{\mathrm{8}\:+\:\frac{\mathrm{7}}{\mathrm{3}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}\:+\:\mathrm{1}}}}\:=\:\mathrm{4}\: \\ $$$$\Rightarrow\:\mathrm{8}\:+\:\frac{\mathrm{7}}{\mathrm{3}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}\:+\:\mathrm{1}}}\:=\:\mathrm{3} \\ $$$$\Rightarrow\:\frac{\mathrm{7}}{\mathrm{3}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}\:+\:\mathrm{1}}}\:=\:−\mathrm{5} \\…
Question Number 146635 by mathdanisur last updated on 14/Jul/21 $$\underset{{x}\rightarrow\infty} {{lim}}\left(\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{{x}\:+\:\mathrm{1}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{x}}}} =\:? \\ $$ Answered by Olaf_Thorendsen last updated on 14/Jul/21 $$\frac{\mathrm{1}}{{x}}\mathrm{ln}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{{x}}\left[\mathrm{ln}\left(\mathrm{2}+{x}\right)−\mathrm{ln}\left({x}+\mathrm{1}\right)\right] \\ $$$$=\:\frac{\mathrm{1}}{{x}}\left[\mathrm{ln2}+\mathrm{ln}{x}+\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)−\mathrm{ln}{x}−\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right] \\…
Question Number 146634 by mathdanisur last updated on 14/Jul/21 $$\boldsymbol{{log}}_{\boldsymbol{{xyz}}} \:\boldsymbol{{x}}\:=\:\mathrm{2}\:\:\:{and}\:\:\:\boldsymbol{{log}}_{\boldsymbol{{xyz}}} \:\boldsymbol{{y}}\:=\:\mathrm{3} \\ $$$${find}\:\:\:\boldsymbol{{log}}_{\boldsymbol{{xyz}}} \:\boldsymbol{{z}}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 14/Jul/21 $$\mathrm{log}_{{xyz}}…
Question Number 146622 by mathdanisur last updated on 14/Jul/21 $$\frac{\mathrm{1}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}\:+\:\mathrm{1}}\:+\:\mathrm{1}\:=\:? \\ $$ Commented by hknkrc46 last updated on 14/Jul/21 $$\bigstar\:\frac{\mathrm{1}}{\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\:\mathrm{1}\right)\left(\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\:\mathrm{1}\right)\left(\sqrt{\mathrm{6}}\:+\:\mathrm{1}\right)}\:+\:\mathrm{1} \\ $$$$=\:\frac{\sqrt[{\mathrm{8}}]{\mathrm{6}}\:−\:\mathrm{1}}{\underset{\sqrt[{\mathrm{4}}]{\mathrm{6}}\:−\:\mathrm{1}} {\underbrace{\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}\:−\:\mathrm{1}\right)\left(\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\:\mathrm{1}\right)}}\left(\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\:\mathrm{1}\right)\left(\sqrt{\mathrm{6}}\:+\:\mathrm{1}\right)}\:+\:\mathrm{1} \\ $$$$=\:\frac{\sqrt[{\mathrm{8}}]{\mathrm{6}}\:−\:\mathrm{1}}{\underset{\sqrt{\mathrm{6}}\:−\:\mathrm{1}}…
Question Number 146608 by mathdanisur last updated on 14/Jul/21 $$\frac{{x}\:\sqrt{{x}}\:-\:\mathrm{1}}{{x}\:+\:\sqrt{{x}}\:+\:\mathrm{1}}\::\:\frac{\sqrt{{x}}\:-\:\mathrm{1}}{\:\sqrt{{x}}\:+\:\mathrm{1}}\:=\:? \\ $$ Answered by qaz last updated on 15/Jul/21 $$\frac{\mathrm{x}\sqrt{\mathrm{x}}−\mathrm{1}}{\mathrm{x}+\sqrt{\mathrm{x}}+\mathrm{1}}:\frac{\sqrt{\mathrm{x}}−\mathrm{1}}{\:\sqrt{\mathrm{x}}+\mathrm{1}}=\frac{\mathrm{a}^{\mathrm{3}} −\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{a}+\mathrm{1}}:\frac{\mathrm{a}−\mathrm{1}}{\mathrm{a}+\mathrm{1}} \\ $$$$=\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{a}+\mathrm{1}}{\mathrm{a}^{\mathrm{2}}…
Question Number 146611 by mathdanisur last updated on 14/Jul/21 $${y}={x}^{\mathrm{2}} +{x}\:,\:{y}=\mathrm{2}{x}^{\mathrm{2}} +{x}+{m}\:\:{and}\:\:{y}={kx}+{b} \\ $$$${if}\:{one}\:{point}\:{intersects},\:{find}\:\:\:{k}\centerdot{b}=? \\ $$ Answered by mr W last updated on 14/Jul/21 $${y}={x}^{\mathrm{2}}…
Question Number 146610 by mathdanisur last updated on 14/Jul/21 $${How}\:{many}\:{roots}\:{does}\:{the}\:{equation} \\ $$$${have}? \\ $$$$\mathrm{5}^{\mathrm{3}\boldsymbol{{x}}} \:+\:\mathrm{27}\centerdot\mathrm{5}^{−\mathrm{3}\boldsymbol{{x}}} \:+\:\mathrm{9}\centerdot\mathrm{5}^{\boldsymbol{{x}}} \:+\:\mathrm{27}\centerdot\mathrm{5}^{−\boldsymbol{{x}}} \:=\:\mathrm{64} \\ $$ Answered by iloveisrael last updated…
Question Number 146606 by iloveisrael last updated on 14/Jul/21 Answered by som(math1967) last updated on 14/Jul/21 $${x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{20} \\ $$$$\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{20} \\…
Question Number 81054 by peter frank last updated on 09/Feb/20 $${solve}\:{the}\:{differential} \\ $$$${equation} \\ $$$$\left({a}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{sin}\:^{\mathrm{2}} \left({x}+{y}\right) \\ $$$$\left({b}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{cos}\:\left({x}+{y}\right) \\ $$ Commented…