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Question-146316




Question Number 146316 by mathdanisur last updated on 12/Jul/21
Answered by Ar Brandon last updated on 12/Jul/21
log_(x−2) (2x+7)≤1  Conditions;  2x+7>0 ∧ 1≠x−2>0,  ⇒(x>−(7/2)) ∧( 0<x−2<1∪x−2>1)  ⇒(x>−(7/2))∧(2<x<3∪x>3)  Case 1: 2<x<3  log_(x−2) (2x+7)≤1⇒2x+7≥x−2⇒x≥−9  ⇒x∈(2, 3) by hypothesis.  Case 2: x>3  log_(x−2) (2x+7)≤1 ⇒2x+7≤x−2⇒x≤−9  Absurd, since x>3 by hypothesis  Unique solution is 2<x<3
$$\mathrm{log}_{\mathrm{x}−\mathrm{2}} \left(\mathrm{2x}+\mathrm{7}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{Conditions}; \\ $$$$\mathrm{2x}+\mathrm{7}>\mathrm{0}\:\wedge\:\mathrm{1}\neq\mathrm{x}−\mathrm{2}>\mathrm{0}, \\ $$$$\Rightarrow\left(\mathrm{x}>−\frac{\mathrm{7}}{\mathrm{2}}\right)\:\wedge\left(\:\mathrm{0}<\mathrm{x}−\mathrm{2}<\mathrm{1}\cup\mathrm{x}−\mathrm{2}>\mathrm{1}\right) \\ $$$$\Rightarrow\left(\mathrm{x}>−\frac{\mathrm{7}}{\mathrm{2}}\right)\wedge\left(\mathrm{2}<\mathrm{x}<\mathrm{3}\cup\mathrm{x}>\mathrm{3}\right) \\ $$$$\mathrm{Case}\:\mathrm{1}:\:\mathrm{2}<\mathrm{x}<\mathrm{3} \\ $$$$\mathrm{log}_{\mathrm{x}−\mathrm{2}} \left(\mathrm{2x}+\mathrm{7}\right)\leqslant\mathrm{1}\Rightarrow\mathrm{2x}+\mathrm{7}\geqslant\mathrm{x}−\mathrm{2}\Rightarrow\mathrm{x}\geqslant−\mathrm{9} \\ $$$$\Rightarrow\mathrm{x}\in\left(\mathrm{2},\:\mathrm{3}\right)\:\mathrm{by}\:\mathrm{hypothesis}. \\ $$$$\mathrm{Case}\:\mathrm{2}:\:\mathrm{x}>\mathrm{3} \\ $$$$\mathrm{log}_{\mathrm{x}−\mathrm{2}} \left(\mathrm{2x}+\mathrm{7}\right)\leqslant\mathrm{1}\:\Rightarrow\mathrm{2x}+\mathrm{7}\leqslant\mathrm{x}−\mathrm{2}\Rightarrow\mathrm{x}\leqslant−\mathrm{9} \\ $$$$\mathrm{Absurd},\:\mathrm{since}\:\mathrm{x}>\mathrm{3}\:\mathrm{by}\:\mathrm{hypothesis} \\ $$$$\mathrm{Unique}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{2}<\mathrm{x}<\mathrm{3} \\ $$
Commented by mathdanisur last updated on 12/Jul/21
Thank you Ser, answer x∈(2;3).?
$${Thank}\:{you}\:{Ser},\:{answer}\:{x}\in\left(\mathrm{2};\mathrm{3}\right).? \\ $$
Commented by Ar Brandon last updated on 12/Jul/21
Yes  x∈(2, 3)⇔2<x<3 ⇔ x∈]2, 3[
$$\mathrm{Yes} \\ $$$$\left.\mathrm{x}\in\left(\mathrm{2},\:\mathrm{3}\right)\Leftrightarrow\mathrm{2}<\mathrm{x}<\mathrm{3}\:\Leftrightarrow\:\mathrm{x}\in\right]\mathrm{2},\:\mathrm{3}\left[\right. \\ $$
Commented by mathdanisur last updated on 12/Jul/21
cool thanks Ser
$${cool}\:{thanks}\:{Ser} \\ $$
Answered by iloveisrael last updated on 13/Jul/21
 log _(x−2) (2x+7)≤log _(x−2) (x−2)  ⇒(x−2−1)(2x+7−x+2)≤0  ⇒(x−3)(x+9)≤0   ⇒−9≤x≤3    (i)  (ii) 2x+7>0⇒x>−(7/2)  (iii)x−2>0⇒x>2  (iv)x−2≠1⇒x≠3  solution : (i)∩(ii)∩(iii)∩(iv)  ⇒ 2<x<3
$$\:\mathrm{log}\:_{\mathrm{x}−\mathrm{2}} \left(\mathrm{2x}+\mathrm{7}\right)\leqslant\mathrm{log}\:_{\mathrm{x}−\mathrm{2}} \left(\mathrm{x}−\mathrm{2}\right) \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{2}−\mathrm{1}\right)\left(\mathrm{2x}+\mathrm{7}−\mathrm{x}+\mathrm{2}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}+\mathrm{9}\right)\leqslant\mathrm{0}\: \\ $$$$\Rightarrow−\mathrm{9}\leqslant\mathrm{x}\leqslant\mathrm{3}\:\:\:\:\left(\mathrm{i}\right) \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{2x}+\mathrm{7}>\mathrm{0}\Rightarrow\mathrm{x}>−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\left(\mathrm{iii}\right)\mathrm{x}−\mathrm{2}>\mathrm{0}\Rightarrow\mathrm{x}>\mathrm{2} \\ $$$$\left(\mathrm{iv}\right)\mathrm{x}−\mathrm{2}\neq\mathrm{1}\Rightarrow\mathrm{x}\neq\mathrm{3} \\ $$$$\mathrm{solution}\::\:\left(\mathrm{i}\right)\cap\left(\mathrm{ii}\right)\cap\left(\mathrm{iii}\right)\cap\left(\mathrm{iv}\right) \\ $$$$\Rightarrow\:\mathrm{2}<\mathrm{x}<\mathrm{3} \\ $$
Commented by mathdanisur last updated on 13/Jul/21
thanks Ser
$${thanks}\:{Ser} \\ $$

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