Question Number 193656 by York12 last updated on 17/Jun/23 $${please}\:{can}\:{someone}\:{solves}\:{this}\: \\ $$ Commented by York12 last updated on 17/Jun/23 Commented by mahdipoor last updated on…
Question Number 193360 by MATHEMATICSAM last updated on 11/Jun/23 $$\mathrm{If}\:{x}\:=\:\mathrm{2}^{{p}} \:\mathrm{and}\:{y}\:=\:\mathrm{4}^{{q}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}^{\mathrm{3}} {y}\right)\:=\:\mathrm{3}{p}\:+\:\mathrm{2}{q} \\ $$ Answered by AST last updated on 11/Jun/23…
Question Number 193295 by MATHEMATICSAM last updated on 09/Jun/23 $$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{{x}}\:\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}} \\ $$$$\boldsymbol{\mathrm{equations}}: \\ $$$$\mathrm{4}^{\frac{{x}}{{y}}\:+\:\frac{{y}}{{x}}} \:=\:\mathrm{32} \\ $$$$\mathrm{log}_{\mathrm{3}} \left({x}\:−\:{y}\right)\:+\:\mathrm{log}_{\mathrm{3}} \left({x}\:+\:{y}\right)\:=\:\mathrm{1} \\ $$ Answered by BaliramKumar last…
Question Number 193253 by MATHEMATICSAM last updated on 08/Jun/23 $$\boldsymbol{\mathrm{Select}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{option}}\:\boldsymbol{\mathrm{with}}\: \\ $$$$\boldsymbol{\mathrm{explaination}}: \\ $$$$\mathrm{If}\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}_{\mathrm{3}} {M}\:+\:\mathrm{3log}_{\mathrm{3}} {N}\:=\:\mathrm{1}\:+\:\mathrm{log}_{\mathrm{0}.\mathrm{008}} \mathrm{5}\:\mathrm{then} \\ $$$$\mathrm{a}.\:{M}^{\mathrm{9}} \:=\:\frac{\mathrm{9}}{{N}} \\ $$$$\mathrm{b}.\:{N}^{\mathrm{9}} \:=\:\frac{\mathrm{9}}{{M}} \\ $$$$\mathrm{c}.\:{M}^{\mathrm{3}}…
Question Number 193230 by MATHEMATICSAM last updated on 07/Jun/23 $$\boldsymbol{\mathrm{Choose}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{option}}: \\ $$$$\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{consecutive}\:\mathrm{positive} \\ $$$$\mathrm{integers}\:\mathrm{and}\:\mathrm{log}\left(\mathrm{1}\:+\:{ac}\right)\:=\:\mathrm{2}{k}\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{is}: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{log}\:{a} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{log}\:{b} \\ $$$$\left.\mathrm{c}\right)\:\mathrm{2} \\ $$$$\left.\mathrm{d}\right)\:\mathrm{1} \\…
Question Number 193183 by York12 last updated on 06/Jun/23 $$ \\ $$$${There}\:{exists}\:{a}\:{unique}\:{positive}\:{integer}\:{a}\:{for} \\ $$$${which}\:{The}\:{sum}\:{u}\:\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\lfloor\frac{{n}^{\mathrm{2}} −{na}}{\mathrm{5}}\rfloor\:{is}\:{an}\:{integer} \\ $$$${strictly}\:{between}\:−\mathrm{1000}\:\&\:\mathrm{1000}\:{find}\:{a}+{u}. \\ $$ Answered by York12 last…
Question Number 65492 by Masumsiddiqui399@gmail.com last updated on 30/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 65333 by aliesam last updated on 28/Jul/19 Commented by Prithwish sen last updated on 28/Jul/19 $$\mathrm{log}_{\mathrm{y}} \mathrm{x}=\mathrm{v}\Rightarrow\frac{\mathrm{v}^{\mathrm{2}} +\mathrm{1}}{\mathrm{v}}\:=\:\frac{\mathrm{17}}{\mathrm{4}}\Rightarrow\mathrm{4v}^{\mathrm{2}} −\mathrm{17v}+\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{4v}^{\mathrm{2}} −\mathrm{16v}−\mathrm{v}+\mathrm{4}=\mathrm{0}\Rightarrow\mathrm{4v}\left(\mathrm{v}−\mathrm{4}\right)−\mathrm{1}\left(\mathrm{v}−\mathrm{4}\right)=\mathrm{0} \\…
Question Number 130858 by mathlove last updated on 29/Jan/21 $$\mathrm{log}\:\left[\sqrt{\boldsymbol{{mx}}^{\mathrm{2}} +\mathrm{3}}\:+\boldsymbol{{x}}\right]=? \\ $$The following subject is a node subject with a value of m Terms…
Question Number 130553 by Study last updated on 26/Jan/21 $${log}\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+….\right)=? \\ $$ Answered by Dwaipayan Shikari last updated on 26/Jan/21 $${log}\left(−{log}\left(\mathrm{1}−{x}\right)\right)={log}\left({log}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\right) \\…