Question Number 18306 by mondodotto@gmail.com last updated on 18/Jul/17 Answered by diofanto last updated on 18/Jul/17 $$\frac{\mathrm{log}\:\mathrm{2}^{\mathrm{4}} \:−\:\mathrm{log}\:\mathrm{3}^{\mathrm{4}} }{\mathrm{log}\:\mathrm{3}^{\mathrm{3}} \:+\:\mathrm{log}\:\mathrm{2}^{\mathrm{3}} }\:=\:\frac{\mathrm{4}\:\mathrm{log}\:\mathrm{2}\:−\:\mathrm{4}\:\mathrm{log}\:\mathrm{3}}{\mathrm{3}\:\mathrm{log}\:\mathrm{3}\:+\:\mathrm{3}\:\mathrm{log}\:\mathrm{2}}\:= \\ $$$$=\:\frac{\mathrm{4}\left(\mathrm{log}\:\mathrm{2}\:−\:\mathrm{log}\:\mathrm{3}\right)}{\mathrm{3}\left(\mathrm{log}\:\mathrm{2}\:+\:\mathrm{log}\:\mathrm{3}\right)}\:=\:\frac{\mathrm{4}\:\mathrm{log}\:\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{3}\:\mathrm{log}\:\mathrm{6}}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\mathrm{log}_{\mathrm{6}} \:\left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\…
Question Number 149378 by fotosy2k last updated on 05/Aug/21 Answered by Ar Brandon last updated on 05/Aug/21 $${S}=\psi\left(\mathrm{4}\right)−\psi\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}+\psi\left(\mathrm{1}\right)−\psi\left(\mathrm{1}\right)=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$${S}=\left(\mathrm{1}−\cancel{\frac{\mathrm{1}}{\mathrm{4}}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\cancel{\frac{\mathrm{1}}{\mathrm{5}}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\cancel{\frac{\mathrm{1}}{\mathrm{6}}}\right)+\left(\cancel{\frac{\mathrm{1}}{\mathrm{4}}}−\cancel{\frac{\mathrm{1}}{\mathrm{7}}}\right)+\centerdot\centerdot\centerdot+ \\ $$$$\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$ Answered…
Question Number 149373 by fotosy2k last updated on 05/Aug/21 Answered by john_santu last updated on 05/Aug/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{\mathrm{3x}}{\mathrm{2}}\right)−\left(\mathrm{1}+\frac{\mathrm{2x}}{\mathrm{2}}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{2x}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{x}}{\mathrm{2}}}{\mathrm{x}}\:×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2x}}=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented…
Question Number 149372 by fotosy2k last updated on 05/Aug/21 Answered by john_santu last updated on 05/Aug/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\mathrm{x}+\mathrm{2sin}\:\mathrm{x}\right)}{\mathrm{sin}\:\mathrm{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{sin}\:\left(\mathrm{x}+\mathrm{2sin}\:\mathrm{x}\right)}{\mathrm{x}+\mathrm{2sin}\:\mathrm{x}}\:.\:\mathrm{x}+\mathrm{2sin}\:\mathrm{x}}{\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}.\:\mathrm{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}+\mathrm{2sin}\:\mathrm{x}}{\mathrm{x}}\:=\:\mathrm{1}+\mathrm{2}=\mathrm{3} \\…
Question Number 83834 by Power last updated on 06/Mar/20 Commented by niroj last updated on 06/Mar/20 $$\:\:\int\:\frac{\:\:\mathrm{1}}{\:\sqrt{\mathrm{x}\left(\mathrm{a}−\mathrm{x}\right)}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{ax}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}=\:\int\:\frac{\:\mathrm{1}}{\:\sqrt{−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}.\frac{\mathrm{a}}{\mathrm{2}}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\right)}}\mathrm{dx} \\ $$$$\:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{−\left[\left(\mathrm{x}−\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 18299 by tawa tawa last updated on 18/Jul/17 $$\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \:=\:\mathrm{2},\:\:\:\:\:\:\mathrm{find}\:\:\mathrm{x} \\ $$ Commented by mrW1 last updated on 19/Jul/17 $$\mathrm{no}\:\mathrm{analytic}\:\mathrm{solution} \\ $$$$\sqrt{\mathrm{2}}<\mathrm{x}<\mathrm{2}…
Question Number 18290 by 18±1 last updated on 18/Jul/17 Commented by 1kanika# last updated on 18/Jul/17 $${ans}.\:{is}\:{B}. \\ $$ Commented by 1kanika# last updated on…
Question Number 83824 by Power last updated on 06/Mar/20 Commented by mathmax by abdo last updated on 06/Mar/20 $${changement}\:{x}=\frac{\pi}{\mathrm{2}}−{t}\:{give} \\ $$$$\Omega\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\pi\left(\frac{\pi}{\mathrm{2}}−{t}\right)+{cos}^{\mathrm{4}} {t}−{sin}^{\mathrm{4}} {t}}{{cos}^{\mathrm{4}}…
Question Number 83822 by Power last updated on 06/Mar/20 Commented by mathmax by abdo last updated on 07/Mar/20 $${I}\:=\int\:\:\frac{{dx}}{\:\sqrt{\mathrm{2}{x}−\mathrm{1}}−^{\mathrm{4}} \sqrt{\mathrm{2}{x}−\mathrm{1}}}\:{cha}\mathrm{7}{gement}\:\:^{\mathrm{4}} \sqrt{\mathrm{2}{x}−\mathrm{1}}={t}\:{give} \\ $$$$\mathrm{2}{x}−\mathrm{1}\:={t}^{\mathrm{4}} \:\Rightarrow\mathrm{2}{x}=\mathrm{1}+{t}^{\mathrm{4}} \:\Rightarrow\:\mathrm{2}{dx}\:=\mathrm{4}{t}^{\mathrm{3}}…
Question Number 18285 by chux last updated on 18/Jul/17 $$\int_{\mathrm{0}} ^{\mathrm{5}} \frac{\mathrm{1}}{\int_{\mathrm{1}} ^{\mathrm{8}} \mathrm{e}^{\mathrm{x}^{−\mathrm{5}} } }\mathrm{dx} \\ $$ Commented by chux last updated on 18/Jul/17…