Question Number 159507 by cortano last updated on 18/Nov/21 Commented by tounghoungko last updated on 18/Nov/21 $$\:\frac{\mathrm{5}.\mathrm{5}^{−\mathrm{2}{x}} +\mathrm{5}^{\mathrm{3}} .\mathrm{5}^{−\mathrm{2}{x}} }{\frac{\mathrm{5}}{\mathrm{8}}.\mathrm{2}^{\mathrm{2}{x}} }\:=\:\mathrm{4} \\ $$$$\Leftrightarrow\:\mathrm{5}^{−\mathrm{2}{x}} +\mathrm{5}^{\mathrm{2}} .\mathrm{5}^{−\mathrm{2}{x}}…
Question Number 93957 by O Predador last updated on 16/May/20 $$\: \\ $$$$\:\mathrm{log}_{\sqrt{\mathrm{17}}−\sqrt{\mathrm{2}}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{19}+\sqrt{\mathrm{136}}}}\right)\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{log}_{\sqrt{\mathrm{19}}−\sqrt{\mathrm{3}}} \left(\frac{\mathrm{1}}{\mathrm{22}−\sqrt{\mathrm{228}}}\right)\mathrm{x}\:=\:\mathrm{3} \\ $$$$\: \\ $$$$\:\mathrm{x}\:=\:? \\ $$ Commented by PRITHWISH…
Question Number 93847 by i jagooll last updated on 15/May/20 $$\mathrm{log}\:_{\mathrm{5}} \:\left(\mathrm{x}−\mathrm{2}\right)+\mathrm{log}\:_{\mathrm{8}} \:\left(\mathrm{x}−\mathrm{4}\right)=\:\mathrm{log}\:_{\mathrm{6}} \left(\mathrm{x}−\mathrm{1}\right) \\ $$ Commented by john santu last updated on 15/May/20 $$…
Question Number 159157 by physicstutes last updated on 13/Nov/21 Answered by mr W last updated on 13/Nov/21 Commented by mr W last updated on 13/Nov/21…
Question Number 159122 by physicstutes last updated on 13/Nov/21 $$\mathrm{In}\:\mathrm{order}\:\mathrm{to}\:\mathrm{monitor}\:\mathrm{buses}\:\mathrm{in}\:\mathrm{a}\:\mathrm{travel} \\ $$$$\mathrm{agency},\:\mathrm{the}\:\mathrm{manager}\:\mathrm{decides}\:\mathrm{to}\:\mathrm{monitor} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{break}\:\mathrm{downs}\:\mathrm{of}\:\mathrm{the}\:\mathrm{buses} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{sequence}\:\left\{{x}_{{n}} \right\}\:\mathrm{defined}\:\mathrm{by} \\ $$$${x}_{{n}+\mathrm{1}} \:=\:\mathrm{1}.\mathrm{05}\:{x}_{{n}} \:+\:\mathrm{4}.\:\mathrm{Given}\:\mathrm{that}\:{x}_{\mathrm{0}} \:=\:\mathrm{40}. \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{break}\:\mathrm{downs}\:\mathrm{by}\:\mathrm{the}\:\mathrm{buses} \\…
Question Number 93546 by allizzwell23 last updated on 13/May/20 $$\:\:\mathrm{Which}\:\mathrm{app}\:\mathrm{is}\:\mathrm{the}\:\mathrm{best}\:\mathrm{to}\:\mathrm{evaluate} \\ $$$$\:\:\:\:\frac{\mathrm{W}\left(\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}} \\ $$ Commented by IbrahimGR376 last updated on 13/May/20 $$\:\:\:\mathrm{mathematica}\:\mathrm{app}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{GR376} \\…
Question Number 93428 by mhmd last updated on 13/May/20 Answered by john santu last updated on 13/May/20 $$\mathrm{10}−\mathrm{x}=\mathrm{2}\mid\mathrm{x}\mid\:\Rightarrow\mathrm{100}−\mathrm{20x}+\mathrm{x}^{\mathrm{2}} =\mathrm{4x}^{\mathrm{2}} \\ $$$$\mathrm{3x}^{\mathrm{2}} +\mathrm{20x}−\mathrm{100}=\mathrm{0} \\ $$$$\left(\mathrm{3x}−\mathrm{10}\right)\left(\mathrm{x}+\mathrm{10}\right)=\mathrm{0} \\…
Question Number 92987 by i jagooll last updated on 10/May/20 $$\begin{cases}{\sqrt{\mathrm{y}}\:+\:\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} \right)=\mathrm{2}}\\{\mathrm{y}\:+\:\mathrm{4}\:\mathrm{ln}\left(\mathrm{x}\right)\:=\mathrm{28}}\end{cases} \\ $$ Answered by john santu last updated on 10/May/20 $$\mathrm{2ln}\left(\mathrm{x}\right)\:=\:\mathrm{14}−\frac{\mathrm{y}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2ln}\left(\mathrm{x}\right)\:=\:\mathrm{2}−\sqrt{\mathrm{y}}\:…
Question Number 92980 by john santu last updated on 10/May/20 $$\mathrm{log}\:_{\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}{x}}} \:\left(\sqrt{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}}\right)\:+\:\mathrm{log}\:_{\sqrt[{\mathrm{3}\:\:}]{\mathrm{3}{x}^{\mathrm{2}} }} \left(\sqrt{\mathrm{27}{x}}\right)\:\leqslant\:\mathrm{3} \\ $$ Answered by frc2crc last updated on 10/May/20 $$\frac{\mathrm{1}}{\mathrm{9}}<{x}<\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{or}\:\mathrm{3}^{\mathrm{1}/\mathrm{3}}…
Question Number 27321 by 803jaideep@gmail.com last updated on 05/Jan/18 Commented by 803jaideep@gmail.com last updated on 05/Jan/18 $$\mathrm{plz}\:\mathrm{solve}\:\mathrm{it} \\ $$ Answered by mrW1 last updated on…