Menu Close

Category: None

Question-158884

Question Number 158884 by akolade last updated on 10/Nov/21 Answered by ghimisi last updated on 10/Nov/21 $${log}_{\mathrm{3}} \left({a}+\mathrm{1}\right)={log}_{\mathrm{4}} \left({a}+\mathrm{8}\right)={t}\Rightarrow \\ $$$${a}+\mathrm{1}=\mathrm{3}^{{t}} \\ $$$${a}+\mathrm{8}=\mathrm{4}^{{t}} \Rightarrow\mathrm{4}^{{t}} −\mathrm{3}^{{t}}…

Question-158883

Question Number 158883 by akolade last updated on 10/Nov/21 Commented by Rasheed.Sindhi last updated on 10/Nov/21 $$\left(\underset{{a}} {\underbrace{\sqrt{\mathrm{3}}\:+{x}}}\right)^{\mathrm{12}} +\left(\underset{{b}} {\underbrace{\sqrt{\mathrm{3}}\:−{x}}}\right)^{\mathrm{12}} =\mathrm{172928};{x}=? \\ $$$${a}+{b}=\mathrm{2}\sqrt{\mathrm{3}}\:;{ab}=\mathrm{3}−{x}^{\mathrm{2}} ={y}\:\left({say}\right) \\…

If-n-gt-2-and-n-N-prove-1-2-2-1-3-2-1-n-2-lt-n-1-1-2-1-1-n-1-n-

Question Number 93315 by Shakhzod last updated on 12/May/20 $${If}\:\:{n}>=\mathrm{2}\:\:{and}\:{n}\subset\mathbb{N}\:. \\ $$$${prove}\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…\frac{\mathrm{1}}{{n}^{\mathrm{2}} }<={n}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right)^{\frac{\mathrm{1}}{{n}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

tan-x-dx-Who-can-solve-this-problem-

Question Number 93304 by Shakhzod last updated on 12/May/20 $$\int\sqrt{\mathrm{tan}\:\left({x}\right)}{dx}\:{Who}\:{can}\:{solve}\:{this}\:{problem}? \\ $$ Commented by john santu last updated on 12/May/20 $$=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}+\sqrt{\mathrm{cot}\:\mathrm{x}}}{\:\sqrt{\mathrm{2}}}\right)\:+\: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{cot}\:\mathrm{x}}}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{c}\:\:…

Question-27718

Question Number 27718 by mondodotto@gmail.com last updated on 13/Jan/18 Answered by prakash jain last updated on 13/Jan/18 $$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} ={a} \\ $$$$\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} =\frac{\mathrm{1}}{{a}} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{4}\Rightarrow{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1}=\mathrm{0}…