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find-x-log-log2-log-2-x-1-0-

Question Number 35743 by mondodotto@gmail.com last updated on 22/May/18 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{{x}} \\ $$$$\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{log}}\mathrm{2}+\boldsymbol{\mathrm{log}}_{\mathrm{2}} \left(\boldsymbol{{x}}+\mathrm{1}\right)\right)=\mathrm{0} \\ $$ Commented by prof Abdo imad last updated on 25/May/18 $${if}\:{log}\:{mean}\:{ln}\:\:\left({e}\right)\Leftrightarrow\:{ln}\left(\mathrm{2}\right)\:+{ln}_{\mathrm{2}}…

Did-I-miss-some-updates-Do-we-get-a-prize-or-at-least-an-award-for-the-fastest-answer-Or-for-the-best-or-for-the-most-sophisticated-answer-Or-for-using-the-largest-font-size-and-the-brightest-co

Question Number 101278 by MJS last updated on 01/Jul/20 $$\mathrm{Did}\:\mathrm{I}\:\mathrm{miss}\:\mathrm{some}\:\mathrm{updates}? \\ $$$$\mathrm{Do}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{prize}\:\mathrm{or}\:\mathrm{at}\:\mathrm{least}\:\mathrm{an}\:\mathrm{award}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{fastest}\:\mathrm{answer}?\:\mathrm{Or}\:\mathrm{for}\:\mathrm{the}\:“\mathrm{best}'',\:\mathrm{or}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{most}\:\mathrm{sophisticated}\:\mathrm{answer}?\:\mathrm{Or}\:\mathrm{for}\:\mathrm{using}\:\mathrm{the} \\ $$$$\mathrm{largest}\:\mathrm{font}\:\mathrm{size}\:\mathrm{and}\:\mathrm{the}\:\mathrm{brightest}\:\mathrm{colour}? \\ $$$$\mathrm{Annoying}\:\mathrm{developments}… \\ $$ Commented by MJS…

f-x-x-1-x-show-that-K-R-such-that-x-y-R-f-x-f-y-K-x-y-

Question Number 166805 by mathocean1 last updated on 28/Feb/22 $${f}\left({x}\right)=\frac{{x}}{\mathrm{1}+\mid{x}\mid}. \\ $$$${show}\:{that}\:\exists\:{K}\:\in\:\mathbb{R}_{+} \:{such}\:{that} \\ $$$$\forall\:{x},\:{y}\:\in\:\mathbb{R},\:\mid{f}\left({x}\right)−{f}\left({y}\right)\mid\leqslant{K}\mid{x}−{y}\mid \\ $$ Answered by TheSupreme last updated on 28/Feb/22 $${f}'\left({x}\right)=\frac{\mathrm{1}+\mid{x}\mid−\mid{x}\mid}{\left(\mathrm{1}+\mid{x}\mid\right)^{\mathrm{2}}…

Question-166793

Question Number 166793 by leicianocosta last updated on 28/Feb/22 Answered by Rasheed.Sindhi last updated on 28/Feb/22 $$\mathrm{r}=\mathrm{a}+\mathrm{2}=\mathrm{b}+\mathrm{9} \\ $$$$\mathrm{a}=\mathrm{r}−\mathrm{2}\:,\:\mathrm{b}=\mathrm{r}−\mathrm{9} \\ $$$$\mathrm{diagonal}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \\…