Question Number 146725 by mathdanisur last updated on 15/Jul/21 $${cos}\left({x}\right)\:\centerdot\:{cos}\left(\mathrm{3}{x}\right)\:=\:{cos}\left(\mathrm{5}{x}\right)\:\centerdot\:{cos}\left(\mathrm{7}{x}\right) \\ $$$$\Rightarrow\:{x}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 15/Jul/21 $$\mathrm{cos}{x}.\mathrm{cos3}{x}\:=\:\mathrm{cos5}{x}.\mathrm{cos7}{x} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\left({x}−\mathrm{3}{x}\right)+\mathrm{cos}\left({x}+\mathrm{3}{x}\right)\right]\:= \\…
Question Number 146680 by mathdanisur last updated on 14/Jul/21 $${Compare}:\:\:\mathrm{100}^{\mathrm{101}} \:\:{and}\:\:\:\mathrm{101}^{\mathrm{100}} \\ $$ Answered by Olaf_Thorendsen last updated on 14/Jul/21 $$\mathrm{log}_{\mathrm{10}} \mathrm{101}\:=\:\mathrm{2}+\mathrm{log}_{\mathrm{10}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{100}}\right) \\ $$$$\mathrm{100log}_{\mathrm{10}}…
Question Number 146672 by mathdanisur last updated on 14/Jul/21 Answered by Olaf_Thorendsen last updated on 15/Jul/21 $$\mathrm{S}\:=\:\frac{\underset{{n}=\mathrm{1}} {\overset{{x}} {\sum}}{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\underset{{n}=\mathrm{1}} {\overset{{x}} {\sum}}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \\ $$$$\mathrm{S}\:=\:\frac{\underset{{n}=\mathrm{1}}…
Question Number 146678 by mathdanisur last updated on 14/Jul/21 $$\int\:\sqrt{\mathrm{2}\:+\:{x}^{\mathrm{2}} }\:{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 14/Jul/21 $$\mathrm{I}=\int\sqrt{\mathrm{2}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:=\mathrm{2}\int\sqrt{\mathrm{1}+\mathrm{sinh}^{\mathrm{2}}…
Question Number 146677 by mathdanisur last updated on 14/Jul/21 $${if}\:\:\:{f}\left({x}\right)\:=\:\mathrm{3}^{\boldsymbol{{x}}+\mathrm{1}} \:\:\:{find}\:\:\:\frac{{f}\left(\mathrm{2}{x}\:+\:\mathrm{1}\right)}{{f}\left({x}\:+\:\mathrm{1}\right)}\:=\:? \\ $$ Answered by hknkrc46 last updated on 14/Jul/21 $$\blacktriangleright\:\boldsymbol{{f}}\left(\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{1}\right)\:=\:\mathrm{3}^{\left(\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{1}\right)\:+\:\mathrm{1}} \:=\:\mathrm{3}^{\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{2}} \\ $$$$\blacktriangleright\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\:+\:\mathrm{1}\right)\:=\:\mathrm{3}^{\left(\boldsymbol{{x}}\:+\:\mathrm{1}\right)\:+\:\mathrm{1}} \:=\:\mathrm{3}^{\boldsymbol{{x}}\:+\:\mathrm{2}}…
Question Number 81139 by TawaTawa last updated on 09/Feb/20 Commented by TawaTawa last updated on 09/Feb/20 $$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}. \\ $$ Commented by mr W last updated…
Question Number 146653 by mathdanisur last updated on 14/Jul/21 Answered by mindispower last updated on 14/Jul/21 $${tg}\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\frac{\mathrm{1}}{{tg}\left({x}\right)} \\ $$$${S}=\underset{{k}=\mathrm{20}} {\overset{\mathrm{70}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)} \\ $$$$=\underset{{k}=\mathrm{20}} {\overset{\mathrm{44}} {\sum}}\frac{{tg}\left({k}\right)}{\mathrm{1}+{tg}\left({k}\right)}+\frac{{tg}\left(\mathrm{45}\right)}{{tg}\left(\mathrm{45}\right)+\mathrm{1}}+\underset{{k}=\mathrm{46}}…
Question Number 15570 by tawa tawa last updated on 11/Jun/17 $$\mathrm{Solve}:\:\:\:\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{x}^{\mathrm{4}} \\ $$ Commented by tawa tawa last updated on 11/Jun/17 $$\mathrm{workings} \\ $$…
Question Number 146636 by mathdanisur last updated on 14/Jul/21 $$\mathrm{4}\:+\:\frac{\mathrm{12}}{\mathrm{8}\:+\:\frac{\mathrm{7}}{\mathrm{3}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}\:+\:\mathrm{1}}}}\:=\:\mathrm{8}\:\:\Rightarrow\:\:\boldsymbol{{x}}=? \\ $$ Commented by hknkrc46 last updated on 14/Jul/21 $$\bigstar\:\frac{\mathrm{12}}{\mathrm{8}\:+\:\frac{\mathrm{7}}{\mathrm{3}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}\:+\:\mathrm{1}}}}\:=\:\mathrm{4}\: \\ $$$$\Rightarrow\:\mathrm{8}\:+\:\frac{\mathrm{7}}{\mathrm{3}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}\:+\:\mathrm{1}}}\:=\:\mathrm{3} \\ $$$$\Rightarrow\:\frac{\mathrm{7}}{\mathrm{3}\:+\:\frac{\mathrm{4}}{\boldsymbol{{x}}\:+\:\mathrm{1}}}\:=\:−\mathrm{5} \\…
Question Number 146635 by mathdanisur last updated on 14/Jul/21 $$\underset{{x}\rightarrow\infty} {{lim}}\left(\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{{x}\:+\:\mathrm{1}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{x}}}} =\:? \\ $$ Answered by Olaf_Thorendsen last updated on 14/Jul/21 $$\frac{\mathrm{1}}{{x}}\mathrm{ln}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{{x}}\left[\mathrm{ln}\left(\mathrm{2}+{x}\right)−\mathrm{ln}\left({x}+\mathrm{1}\right)\right] \\ $$$$=\:\frac{\mathrm{1}}{{x}}\left[\mathrm{ln2}+\mathrm{ln}{x}+\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)−\mathrm{ln}{x}−\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right] \\…