Question Number 185414 by Safiullah_21 last updated on 21/Jan/23 $$\mathrm{2}^{{x}} ={a} \\ $$$$\mathrm{3}^{{y}} ={b}\:\:\:\:\:\::\:\mathrm{6}^{{xy}} =? \\ $$$$\Rightarrow\mathrm{6}^{{xy}} =\left(\mathrm{2}×\mathrm{3}\right)^{{xy}} \Rightarrow\mathrm{2}^{{xy}} ×\mathrm{3}^{{xy}} \Rightarrow\left(\mathrm{2}^{{x}} \right)^{{y}} ×\left(\mathrm{3}^{{y}} \right)^{{x}} \Rightarrow{a}^{{y}}…
Question Number 185403 by mathlove last updated on 21/Jan/23 $$\underset{{k}=\mathrm{1}} {\overset{\mathrm{16}} {\sum}}\left(\sqrt{{k}}−\sqrt{{k}−\mathrm{1}}\right)=? \\ $$ Answered by aba last updated on 21/Jan/23 $$=\mathrm{4} \\ $$ Answered…
Question Number 185395 by mathlove last updated on 21/Jan/23 Answered by witcher3 last updated on 21/Jan/23 $$\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}=\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{2k}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\left(\mathrm{2k}\right)^{\mathrm{2}} +\mathrm{2k}+\frac{\mathrm{1}}{\mathrm{2}} \\…
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Question Number 185394 by mathlove last updated on 21/Jan/23 Answered by mr W last updated on 21/Jan/23 $${P}×\left(\mathrm{3}^{\mathrm{2}^{\mathrm{0}} } −\mathrm{1}\right)=\left(\mathrm{3}^{\mathrm{2}^{\mathrm{1}} } −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{2}^{\mathrm{1}} } +\mathrm{1}\right)…\left(\mathrm{3}^{\mathrm{2}^{{n}} }…
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Question Number 119848 by benjo_mathlover last updated on 27/Oct/20 $${Solve}\:{in}\:{real}\:{numbers}\:{the}\:{equation} \\ $$$$\sqrt[{\mathrm{3}\:}]{{x}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}−\mathrm{1}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}+\mathrm{1}}\:=\:\mathrm{0} \\ $$ Answered by 1549442205PVT last updated on 27/Oct/20 $$\mathrm{Applying}\:\mathrm{the}\:\mathrm{identity}\: \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}}…
Question Number 119849 by benjo_mathlover last updated on 27/Oct/20 $${Find}\:{all}\:{pair}\left({x},{y}\right)\:{of}\:{real}\:{numbers} \\ $$$${that}\:{are}\:{the}\:{solutions}\:{to}\:{the}\:{system} \\ $$$$\begin{cases}{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −{y}=−\frac{\mathrm{1}}{\mathrm{4}}+\sqrt{\mathrm{3}}}\\{{y}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{3}} −{x}=−\frac{\mathrm{1}}{\mathrm{4}}−\sqrt{\mathrm{3}}}\end{cases} \\ $$ Answered by 1549442205PVT last updated…
Question Number 185383 by aba last updated on 20/Jan/23 $$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\sqrt[{\mathrm{n}}]{\left(_{\mathrm{1}} ^{\mathrm{n}} \right)\left(_{\mathrm{2}} ^{\mathrm{n}} \right)…\left(_{\mathrm{n}} ^{\mathrm{n}} \right)}}{\mathrm{e}^{\frac{\mathrm{n}}{\mathrm{2}}} ×\mathrm{n}^{−\frac{\mathrm{1}}{\mathrm{2}}} }=? \\ $$ Terms of Service Privacy…
Question Number 119832 by talminator2856791 last updated on 27/Oct/20 $$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{evaluate}:\:\:\:\:{I}\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}!} {dx} \\ $$$$ \\ $$$$ \\ $$ Terms of…