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Question Number 120927 by Ar Brandon last updated on 04/Nov/20

Answered by 675480065 last updated on 04/Nov/20

$$\mathrm{domain}:2\mathrm{x}-3/4>0\cap \mathrm{x}>0\cap \mathrm{x}>1$$$$\Rightarrow \mathrm{x}>3/8\cap \mathrm{x}>0$$$$\mathrm{hence}2\mathrm{x}-3/4>{\mathrm{x}}^2$$$$\Rightarrow {\mathrm{x}}^2-2\mathrm{x}+3/4<0$$$$\Rightarrow \frac{1}{2}<\mathrm{x}<\frac{3}{2}$$$$\mathrm{Final}\mathrm{ans}:$$$$(\frac{3}{8},\frac{1}{2})\cup (1,\frac{3}{2})$$$$\mathrm{B}$$$$$$

Commented by Ar Brandon last updated on 04/Nov/20

$$\mathrm{Thanks}\mathrm{Sir}.$$$$\mathrm{But}\mathrm{I}\mathrm{got}(\frac{1}{2},1)\cup (1,\frac{3}{2})$$$$\mathrm{Any}\mathrm{opinion},\mathrm{please}?$$

Commented by Ar Brandon last updated on 04/Nov/20

$$\mathrm{Since}\mathrm{for}\mathrm{domaine}\mathrm{we}\mathrm{have}$$$$2\mathrm{x}-3/4>0\wedge \mathrm{x}>0,\mathrm{x}\ne 1$$$$\Rightarrow \mathrm{x}>3/8\wedge \mathrm{x}>0,\mathrm{x}\ne 1$$$$\mathrm{and}\frac{1}{2}<\mathrm{x}<\frac{3}{2}$$

Commented by 675480065 last updated on 04/Nov/20

$$\mathrm{this}\mathrm{is}\mathrm{false}$$$$$$

Answered by liberty last updated on 04/Nov/20

$$\left(1\right)\iff \log {}_{\mathrm{x}}(2\mathrm{x}-\frac{3}{4})>\log {}_{\mathrm{x}}\left({\mathrm{x}}^2\right)$$$$\iff (\mathrm{x}-1)(2\mathrm{x}-\frac{3}{4}-{\mathrm{x}}^2)>0$$$$\Rightarrow (\mathrm{x}-1)({4\mathrm{x}}^2-8\mathrm{x}+3)<0$$$$\Rightarrow (\mathrm{x}-1)(2\mathrm{x}-1)(2\mathrm{x}-3)<0$$$$\Rightarrow \mathrm{x}<\frac{1}{2}\cup 1<\mathrm{x}<\frac{3}{2}$$$$\left(2\right)\iff \mathrm{numerus}>0;2\mathrm{x}>\frac{3}{4};\mathrm{x}>\frac{3}{8}$$$$\left(3\right)\iff \mathrm{base}\mathrm{x}>0$$$$\mathrm{therefore}\mathrm{we}\mathrm{get}\mathrm{solution}\mathrm{from}$$$$\left(1\right)\cap \left(2\right)\cap \left(3\right)\Rightarrow (\frac{3}{8},\frac{1}{2})\cup (1,\frac{3}{2})$$

Commented by Ar Brandon last updated on 04/Nov/20

Thanks for your efforts Sir

Answered by ebi last updated on 04/Nov/20

$$$$$${log}_x(2x-\frac{3}{4})>2$$$$\frac{log(2x-\frac{3}{4})}{logx}>2$$$$$$$$case\left(A\right)$$$$logx>0$$$$\therefore x>1$$$$$$$$case\left(B\right)$$$$2x-\frac{3}{4}>0$$$$\therefore x>\frac{3}{8}$$$$$$$$case\left(C\right)$$$$log(2x-\frac{3}{4})>log{x}^2$$$$2x-\frac{3}{4}>x^2$$$$x^2-2x+\frac{3}{4}<0$$$$(x-\frac{1}{2})(x-\frac{3}{2})<0$$$$\therefore \frac{1}{2}<x<\frac{3}{2}$$$$$$$$solution$$$$\left(A\right)\cap \left(B\right)\cap \left(C\right)$$$$x=(\frac{3}{8},\frac{1}{2})\cup (1,\frac{3}{2})$$

Commented by Ar Brandon last updated on 04/Nov/20

Thank you

Answered by Ar Brandon last updated on 04/Nov/20

$${\log }_{\mathrm{x}}(2\mathrm{x}-\frac{3}{4})>2={\log }_{\mathrm{x}}{\mathrm{x}}^2$$$$\mathrm{Domaine}:2\mathrm{x}-\frac{3}{4}>0\wedge \mathrm{x}>0,\mathrm{x}\ne 1$$$$\Rightarrow \mathrm{x}>\frac{3}{8}\wedge \mathrm{x}>0,\mathrm{x}\ne 1$$$$\Rightarrow \mathrm{Domaine}\mathrm{is}\mathrm{x}\in (\frac{3}{8},1)\cup (1,+\mathrm{\infty })$$$$\mathrm{Case}1:\mathrm{when}\mathrm{x}\in (\frac{3}{8},1)$$$${\log }_{\mathrm{x}}(2\mathrm{x}-\frac{3}{4})>{\log }_{\mathrm{x}}{\mathrm{x}}^2$$$$\Rightarrow 2\mathrm{x}-\frac{3}{4}<{\mathrm{x}}^2,\mathrm{since}\mathrm{function}\mathrm{is}\mathrm{decreasing}\mathrm{for}$$$$\mathrm{x}\in (\frac{3}{8},1)\mathrm{ie}{\log }_{\mathrm{x}}\mathrm{a}>{\log }_{\mathrm{x}}\mathrm{b}\Rightarrow \mathrm{a}<\mathrm{b}$$$$\Rightarrow {4\mathrm{x}}^2-8\mathrm{x}+3>0\Rightarrow (2\mathrm{x}-1)(2\mathrm{x}-3)>0$$$$\Rightarrow \mathrm{x}\in (-\mathrm{\infty },\frac{1}{2})\cup (\frac{3}{2},+\mathrm{\infty })$$$$\mathrm{But}\mathrm{from}\mathrm{initial}\mathrm{condition}\mathrm{x}\in (\frac{3}{8},1)$$$$\mathrm{Therefore}\mathrm{x}\in (\frac{3}{8},\frac{1}{2})$$$$\mathrm{Case}2:\mathrm{when}\mathrm{x}>1$$$${\log }_{\mathrm{x}}(2\mathrm{x}-\frac{3}{4})>{\log }_{\mathrm{x}}{\mathrm{x}}^2$$$$2\mathrm{x}-\frac{3}{4}>{\mathrm{x}}^2,\mathrm{since}\mathrm{function}\mathrm{is}\mathrm{increasing}\mathrm{for}$$$$\mathrm{x}>1.\mathrm{ie}{\log }_{\mathrm{x}}\mathrm{a}>{\log }_{\mathrm{x}}\mathrm{b}\Rightarrow \mathrm{a}>\mathrm{b}$$$$\Rightarrow {4\mathrm{x}}^2-8\mathrm{x}+3<0\Rightarrow (2\mathrm{x}-1)(2\mathrm{x}-3)<0$$$$\Rightarrow \frac{1}{2}<\mathrm{x}<\frac{3}{2}$$$$\mathrm{But}\mathrm{from}\mathrm{initial}\mathrm{condition}\mathrm{x}>1$$$$\mathrm{Therefore}\mathrm{x}\in (1,\frac{3}{2})$$$$\mathrm{Combining}\mathrm{case}1\mathrm{and}\mathrm{case}2\mathrm{we}\mathrm{obtain}$$$$\mathrm{x}\in (\frac{3}{8},\frac{1}{2})\cup (1,\frac{3}{2})$$

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