Question Number 25278 by Tinkutara last updated on 07/Dec/17 $${Find}\:{the}\:{number}\:{of}\:{solutions}\:{of} \\ $$$$\mathrm{log}\mid{x}\mid\:=\:{e}^{{x}} \\ $$ Commented by Tinkutara last updated on 08/Dec/17 $${Can}\:{it}\:{be}\:{solved}\:{using}\:{Lambert}'{s}\:{W} \\ $$$${function}? \\…
Question Number 156338 by MathSh last updated on 10/Oct/21 $$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{1}}}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:-\:\mathrm{1} \\ $$$$ \\ $$ Commented by mr W last updated on 10/Oct/21…
Question Number 156333 by MathSh last updated on 10/Oct/21 Answered by Rasheed.Sindhi last updated on 10/Oct/21 $$\underline{\mathrm{NOT}\:\mathrm{General}\:\mathrm{Solution}} \\ $$$$\mathrm{Simple}\:\mathrm{special}\:\mathrm{case}:\mathrm{n}=\mathrm{1} \\ $$$$\mathrm{x}_{\mathrm{1}} =\sqrt{\mathrm{x}_{\mathrm{1}} +\mathrm{22}}\:−\sqrt{\mathrm{x}_{\mathrm{1}} +\mathrm{1}}\: \\…
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Question Number 156328 by MathSh last updated on 10/Oct/21 Commented by MJS_new last updated on 10/Oct/21 $$\mathrm{it}'\mathrm{s}\:\mathrm{wrong}.\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:>\mathrm{2} \\ $$ Commented by MathSh last updated on…
Question Number 90793 by kelum last updated on 26/Apr/20 $${a}+{b}+{c}+{d}=\mathrm{4} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{10} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} =\mathrm{22} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}}…
Question Number 25246 by ibraheem160 last updated on 06/Dec/17 $$\mathrm{8x}^{\frac{\mathrm{3}}{\mathrm{2n}}} −\mathrm{8x}^{\frac{−\mathrm{3}}{\mathrm{2n}}} \:=\mathrm{63} \\ $$ Answered by Rasheed.Sindhi last updated on 07/Dec/17 $$\mathrm{8x}^{\frac{\mathrm{3}}{\mathrm{2n}}} −\mathrm{8x}^{\frac{−\mathrm{3}}{\mathrm{2n}}} \:=\mathrm{63} \\…
Question Number 156313 by cortano last updated on 10/Oct/21 $$\:\mathrm{rationalize}\:\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\sqrt[{\mathrm{3}}]{\mathrm{6}}−\sqrt[{\mathrm{3}}]{\mathrm{9}}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 90772 by MWSuSon last updated on 26/Apr/20 $${show}\:{that}\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}=\left({b}−{c}\right)^{\mathrm{2}} −\mathrm{1}\:{are}\:{rational}\:{if} \\ $$$${b}\:{and}\:{c}\:{are}\:{rational}\:{numbers}. \\ $$ Commented by jagoll last updated on 26/Apr/20…
Question Number 25226 by Mr eaay last updated on 06/Dec/17 $${Show}\:{that}\:{if}\:{x}=\mathrm{3}−\sqrt{\mathrm{3}}.{Show}\:{that}\:{x}^{\mathrm{2}} +\frac{\mathrm{36}}{{x}^{\mathrm{2}} }=\mathrm{24} \\ $$ Answered by naka3546 last updated on 06/Dec/17 $${x}^{\mathrm{2}} \:\:=\:\:\left(\mathrm{3}\:−\:\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{12}\:−\:\mathrm{6}\sqrt{\mathrm{3}}\:\:=\:\:\mathrm{6}\:\left(\mathrm{2}\:−\:\sqrt{\mathrm{3}}\right)…