Question Number 70196 by smartsmith459@gmail.com last updated on 01/Oct/19 Answered by MJS last updated on 01/Oct/19 $${x}={a}^{\mathrm{log}_{{b}} \:{c}} ={a}^{\frac{\mathrm{ln}\:{c}}{\mathrm{ln}\:{b}}} \\ $$$$\mathrm{ln}\:{x}\:=\frac{\mathrm{ln}\:{c}}{\mathrm{ln}\:{b}}\mathrm{ln}\:{a}\:\Rightarrow\:{x}={a}^{\mathrm{log}_{{b}} \:{c}} ={c}^{\mathrm{log}_{{b}} \:{a}} \\…
Question Number 70197 by smartsmith459@gmail.com last updated on 01/Oct/19 Commented by mr W last updated on 01/Oct/19 $${in}\:{the}\:{first}\:\mathrm{27}\:{days}\:{it}\:{climbs}\:\mathrm{27}{m}. \\ $$$${in}\:{the}\:\mathrm{28}{th}\:{day}\:{it}\:{escapes}\:{from}\:{the} \\ $$$${pit}.\:\Rightarrow{totally}\:\mathrm{28}\:{days} \\ $$ Terms…
Question Number 4655 by Yozzii last updated on 18/Feb/16 Commented by Yozzii last updated on 18/Feb/16 $${How}\:{do}\:{you}\:{do}\:{this}?\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 4653 by FilupSmith last updated on 18/Feb/16 $$\mathrm{Silly}\:\mathrm{question} \\ $$$${which}\:{is}\:{correct}??? \\ $$$$ \\ $$$$\mathrm{ln}\left(\frac{{n}}{{ab}}\right)=\mathrm{ln}\left({n}\right)−\mathrm{ln}\left({ab}\right)=\mathrm{ln}\left({n}\right)−\mathrm{ln}\left({a}\right)+\mathrm{ln}\left({b}\right) \\ $$$${or} \\ $$$$\mathrm{ln}\left(\frac{{n}}{{ab}}\right)=\mathrm{ln}\left({n}\right)−\mathrm{ln}\left({ab}\right)=\mathrm{ln}\left({n}\right)−\mathrm{ln}\left({a}\right)−\mathrm{ln}\left({b}\right) \\ $$ Answered by 123456…
Question Number 4650 by FilupSmith last updated on 18/Feb/16 $$\mathrm{In}\:\mathrm{mathematics}\:\mathrm{I}\:\mathrm{have}\:\mathrm{worked}\:\mathrm{with}: \\ $$$$\Sigma\:\mathrm{and}\:\Pi. \\ $$$${e}.{g}. \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{x}} {\sum}}{k}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\left({x}+\mathrm{1}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{x}} {\prod}}{k}\:=\:{k}! \\ $$$$ \\…
Question Number 135723 by Algoritm last updated on 15/Mar/21 Answered by liberty last updated on 15/Mar/21 $$\left(\mathrm{1}\right)\:\sqrt{{x}+\mathrm{3}}\:+\sqrt[{\mathrm{4}}]{\mathrm{9}−{x}}\:\leqslant\:\sqrt{\mathrm{3}}\:\wedge\:{x}^{\mathrm{2}} −\mathrm{16}>\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\sqrt{{x}+\mathrm{3}}\:+\sqrt[{\mathrm{4}}]{\mathrm{9}−{x}}\:\geqslant\sqrt{\mathrm{3}}\:\wedge\:{x}^{\mathrm{2}} −\mathrm{16}>\mathrm{0} \\ $$$${solution}\::\:\left(\mathrm{1}\right)\cup\left(\mathrm{2}\right) \\ $$$$…
Question Number 4645 by FilupSmith last updated on 18/Feb/16 $${S}=\underset{{i}=\mathrm{1}} {\overset{{x}} {\prod}}\left({i}^{\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{i}+\mathrm{1}\right)\left({x}−{i}+\mathrm{2}\right)} \right),\:\left({x},\:{i}\right)\in\mathbb{Z} \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{S}\in\mathbb{Z} \\ $$ Answered by Yozzii last updated on…
Question Number 135718 by liberty last updated on 15/Mar/21 Answered by dhgt last updated on 04/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135712 by Dwaipayan Shikari last updated on 15/Mar/21 Commented by Dwaipayan Shikari last updated on 15/Mar/21 $${Smaller}\:{circle}\:{has}\:{half}\:{of}\:{the}\:{radius}\:{of}\:{the}\:{larger}\:{circle}. \\ $$$${Prove}\:{that},\:{irrespective}\:{of}\:{any}\:{position}\:{oc}\:{the}\:{smaller}\:{circle}\: \\ $$$${that}\:{blue}\:{point}\:{will}\:{always}\:{stay}\:{on}\:{the}\:{diametre}\:{of}\:{that}\: \\ $$$${larger}\:{circle}…
Question Number 135709 by Algoritm last updated on 15/Mar/21 Commented by mr W last updated on 15/Mar/21 $${x}_{{i}} =−\pi\:{or}\:\pi−\mathrm{1} \\ $$$${x}_{\mathrm{2021}} +\pi=\mathrm{0}\:{or}\:\mathrm{2}\pi−\mathrm{1} \\ $$ Answered…