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Solution-log-8-x-log-4-x-log-2-x-11-1-log-x-8-1-log-x-4-1-log-x-2-11-1-log-x-2-3-1-log-x-2-2-1-log-x-2-11-1-3log-x-2-1-2log-x-2-1-log-x-2-11-1-3-1-2-1-

Question Number 70017 by Shamim last updated on 30/Sep/19 $$\mathrm{Solution}- \\ $$$$\mathrm{log}_{\mathrm{8}} \mathrm{x}+\mathrm{log}_{\mathrm{4}} \mathrm{x}+\mathrm{log}_{\mathrm{2}} \mathrm{x}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{8}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{4}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}^{\mathrm{2}}…

find-all-x-y-Z-such-that-x-0-50-y-0-50-x-y-k-k-0-50-x-x-y-99-100-

Question Number 4443 by 123456 last updated on 27/Jan/16 $$\mathrm{find}\:\mathrm{all}\:{x},{y}\in\mathbb{Z}\:\mathrm{such}\:\mathrm{that} \\ $$$${x}\in\left[\mathrm{0},\mathrm{50}\right] \\ $$$${y}\in\left[\mathrm{0},\mathrm{50}\right] \\ $$$${x}+{y}={k},{k}\in\left[\mathrm{0},\mathrm{50}\right] \\ $$$$\frac{{x}}{{x}+{y}}=\frac{\mathrm{99}}{\mathrm{100}} \\ $$ Answered by RasheedSindhi last updated…

Market-is-slow-nowadays-I-mean-Questioning-Answering-Commenting-is-slow-What-are-the-reasons-Winter-season-Shortage-of-problems-Are-we-not-remained-interested-more-Are-we-tir

Question Number 4440 by Rasheed Soomro last updated on 27/Jan/16 $$\mathrm{Market}\:\mathrm{is}\:\mathrm{slow}\:\mathrm{nowadays}!\: \\ $$$$\mathrm{I}\:\mathrm{mean}\:\mathrm{Questioning}/\mathrm{Answering}/\mathrm{Commenting} \\ $$$$\mathrm{is}\:\mathrm{slow}.\:\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{reasons}? \\ $$$$\:\:^{\bullet} \mathrm{Winter}\:\mathrm{season}? \\ $$$$\:\:^{\bullet} \mathrm{Shortage}\:\mathrm{of}\:\mathrm{problems}? \\ $$$$\:\:^{\bullet} \mathrm{Are}\:\mathrm{we}\:\:\mathrm{not}\:\mathrm{remained}\:\mathrm{interested}\:\mathrm{more}? \\…

Question-135431

Question Number 135431 by 777316 last updated on 13/Mar/21 Answered by SEKRET last updated on 13/Mar/21 $$\boldsymbol{\mathrm{F}}\left(\boldsymbol{\mathrm{a}}\right)=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\mathrm{1}\right)}{\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)}\:\boldsymbol{\mathrm{dx}}\:\:\:\:\:\:\boldsymbol{\mathrm{a}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{F}}\:'\left(\boldsymbol{\mathrm{a}}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{ax}}^{\mathrm{2}}…

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Question Number 4297 by 123456 last updated on 07/Jan/16 $$\mathrm{lets} \\ $$$${f}:\left[\mathrm{0},+\infty\right)\rightarrow\mathbb{R},\forall{x}\geqslant{y}\Rightarrow{f}\left({x}\right)\geqslant{f}\left({y}\right) \\ $$$${g}:\left[\mathrm{0},+\infty\right)\rightarrow\mathbb{R} \\ $$$$\mathrm{if} \\ $$$$\forall{x}\in\left[\mathrm{0},+\infty\right),{f}\left({x}\right)\leqslant{g}\left({x}\right)\leqslant{f}\left(\mathrm{2}{x}\right) \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{L},\mathrm{L}\:\mathrm{is}\:\mathrm{finite} \\ $$$$\mathrm{does} \\ $$$$\underset{{x}\rightarrow+\infty}…