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Question Number 87194 by john santu last updated on 03/Apr/20

$$\mathrm{find}\mathrm{the}\mathrm{solution}\mathrm{of}$$$$\frac{\mid {\log }_2\left(\mathrm{x}\right)+2\mid }{\mathrm{x}-3}<2$$

Commented by TANMAY PANACEA. last updated on 03/Apr/20

$$isit(2+{log}_{2}x)or{log}_2(x+2)$$

Commented by john santu last updated on 03/Apr/20

$$2+{\log }_2\left(\mathrm{x}\right)\mathrm{sir}$$

Commented by john santu last updated on 03/Apr/20

$$\mathrm{i}{\mathrm{can}}^{{}^{\prime }}\mathrm{t}\mathrm{answer}\mathrm{that}$$

Commented by john santu last updated on 03/Apr/20

$$\mathrm{the}\mathrm{choice}\mathrm{answer}\mathrm{is}$$$$\mathrm{a})\mathrm{x}<2^3\mathrm{or}\mathrm{x}>2^8$$$$\mathrm{b})3<\mathrm{x}<8$$$$\mathrm{c})\mathrm{x}<3\mathrm{or}\mathrm{x}>8$$$$\mathrm{d})2^3<\mathrm{x}<2^8$$$$\mathrm{e})0<\mathrm{x}<2^3\mathrm{or}\mathrm{x}>2^8$$

Commented by jagoll last updated on 03/Apr/20

$$\mathrm{may}\mathrm{be}\mathrm{the}\mathrm{question}\mathrm{is}$$$$\mid {\log }_2\left(\frac{\mathrm{x}+2}{\mathrm{x}-3}\right)\mid <2$$

Commented by jagoll last updated on 03/Apr/20

$$-2<{\log }_2\left(\frac{\mathrm{x}+2}{\mathrm{x}-3}\right)<2$$$$2^{-2}<\frac{\mathrm{x}+2}{\mathrm{x}-3}<2^2$$$$\Rightarrow \{\begin{array}{l}\frac{\mathrm{x}+2}{\mathrm{x}-3}>\frac{1}{4}\\ \frac{\mathrm{x}+2}{\mathrm{x}-3}<4\end{array}$$

Answered by TANMAY PANACEA. last updated on 03/Apr/20

$$\frac{\mid 2+{log}_{2}x\mid }{x-3}-2<0$$$$1)x\ne 3andx\ne 0$$$$assumex=2^{k}k>0$$$$\frac{\mid 2+k\mid }{2^k-3}-2$$$$=\frac{2+k}{2^k-3}-2$$$$whenk=1$$$$\frac{2+1}{2-3}-2<0$$$$whenk=2$$$$\frac{2+2}{4-3}-2>0\left(notehere\right)$$$$whenk=3$$$$\frac{2+3}{8-3}-2<0$$$$k=4$$$$\frac{2+4}{16-3}-2<0$$$$$$

Answered by MJS last updated on 03/Apr/20

$$\frac{\mid 2+{\log }_{2}x\mid }{x-3}<2\Rightarrow x\in \mathbb{R}\mathrm{\setminus }\{0,3\}$$$$$$$$\left(1\right)2+{\log }_{2}x⩾0\Rightarrow x⩾\frac{1}{4}$$$$\frac{2+{\log }_{2}x}{x-3}<2$$$$(1.1)\frac{1}{4}⩽x<3\Rightarrow x-3<0$$$$256x>4^x\Rightarrow \approx .00392758<x<\approx 5.18752$$$$\Rightarrow \frac{1}{4}⩽x<3\bullet$$$$(1.2)x>3\Rightarrow x-3>0$$$$256x<4^x\Rightarrow x<\approx .00392758\vee x>\approx 5.28752$$$$\Rightarrow x>\approx 5.28752\bullet$$$$\left(2\right)2+{\log }_{2}x<0\Rightarrow x<\frac{1}{4}\Rightarrow x-3<0$$$$\frac{-(2+{\log }_{2}x)}{x-3}<2\mathrm{always}\mathrm{true}\mathrm{because}\mathrm{lhs}<0$$$$\Rightarrow x<0\vee 0<x<\frac{1}{4}\bullet$$$$$$$$\Rightarrow x<0\vee 0<x<3\vee x>\approx 5.28752$$

Commented by MJS last updated on 03/Apr/20

$$\mathrm{even}\mathrm{if}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{wrong},\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{it}$$

Commented by jagoll last updated on 03/Apr/20

$$\mathrm{but}\mathrm{x}<0\mathrm{not}\mathrm{valid}\mathrm{sir}$$$${\log }_2\left(\mathrm{x}\right)\Rightarrow \mathrm{must}\mathrm{be}\mathrm{x}>0$$

Commented by MJS last updated on 03/Apr/20

$$\mathrm{you}\mathrm{are}\mathrm{right}...\mathrm{but}\mid 2+{\log }_{2}x\mid \in \mathbb{R}\mathrm{\forall }x\ne 0...$$$${\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{look}\mathrm{into}\mathrm{this}\mathrm{again}\mathrm{later}$$

Commented by MJS last updated on 03/Apr/20

$${\log }_{2}x=\frac{\ln x}{\ln 2}$$$$\mathrm{for}x\in \mathbb{C}$$$$\frac{\ln x}{\ln 2}=\frac{\ln \left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)}{\ln 2}=\frac{\ln r+\mathrm{i}\theta }{\ln 2}$$$$\mid \frac{\ln r+\mathrm{i}\theta }{\ln 2}\mid =\frac{\sqrt{{\theta }^2+{\left(\ln r\right)}^2}}{\ln 2}$$$$\mathrm{for}x\in {\mathbb{R}}^{-}:\theta =\pi \Rightarrow \mid 2+{\log }_{2}x\mid ⩾0$$$$$$$$\Rightarrow \mathrm{in}\mathrm{our}\mathrm{case}\mathrm{for}x<0\mathrm{the}\mathrm{equation}\mathrm{is}\mathrm{true}$$

Commented by john santu last updated on 03/Apr/20

$$\mathrm{ooo}\mathrm{if}\mathrm{x}\in \mathbb{C}\mathrm{sir}.$$$$\mathrm{so}\mathrm{the}\mathrm{choice}\mathrm{answer}\mathrm{nothing}\mathrm{correct}$$$$\mathrm{sir}$$

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