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Question Number 136256 by EDWIN88 last updated on 20/Mar/21

log _((x−2)) (10−3x) < 2

$$\mathrm{log}\:_{\left(\mathrm{x}−\mathrm{2}\right)} \left(\mathrm{10}−\mathrm{3x}\right)\:<\:\mathrm{2} \\ $$

Answered by liberty last updated on 20/Mar/21

(1) 10−3x > 0 ; 3x−10<0 x < ((10)/3) (2) log _((x−2)) (10−3x) < log _((x−2)) (x−2)^2 (x−2−1)(10−3x−(x−2)^2 ) < 0 ⇔ (x−3)(10−3x−x^2 +4x−4)< 0 (x−3)(−x^2 +x+6)< 0 ⇔(x−3)(x^2 −x−6)>0 (x−3)(x−3)(x+2)>0 (x−3)^2 (x+2)>0 ⇒−2<x<3 ∪ x>3 (3) x−2>0, x>2 The solution set is (2,3)∪(3,((10)/3))

$$\left(\mathrm{1}\right)\:\mathrm{10}−\mathrm{3}{x}\:>\:\mathrm{0}\:;\:\mathrm{3}{x}−\mathrm{10}<\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\:{x}\:<\:\frac{\mathrm{10}}{\mathrm{3}} \\ $$ $$\left(\mathrm{2}\right)\:\mathrm{log}\:_{\left({x}−\mathrm{2}\right)} \left(\mathrm{10}−\mathrm{3}{x}\right)\:<\:\mathrm{log}\:_{\left({x}−\mathrm{2}\right)} \left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$ $$\:\left({x}−\mathrm{2}−\mathrm{1}\right)\left(\mathrm{10}−\mathrm{3}{x}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \right)\:<\:\mathrm{0} \\ $$ $$\Leftrightarrow\:\left({x}−\mathrm{3}\right)\left(\mathrm{10}−\mathrm{3}{x}−{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{4}\right)<\:\mathrm{0} \\ $$ $$\left({x}−\mathrm{3}\right)\left(−{x}^{\mathrm{2}} +{x}+\mathrm{6}\right)<\:\mathrm{0} \\ $$ $$\Leftrightarrow\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{6}\right)>\mathrm{0} \\ $$ $$\left({x}−\mathrm{3}\right)\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)>\mathrm{0} \\ $$ $$\left({x}−\mathrm{3}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right)>\mathrm{0} \\ $$ $$\Rightarrow−\mathrm{2}<{x}<\mathrm{3}\:\cup\:{x}>\mathrm{3} \\ $$ $$\left(\mathrm{3}\right)\:{x}−\mathrm{2}>\mathrm{0},\:{x}>\mathrm{2} \\ $$ $${The}\:{solution}\:{set}\:{is}\:\left(\mathrm{2},\mathrm{3}\right)\cup\left(\mathrm{3},\frac{\mathrm{10}}{\mathrm{3}}\right) \\ $$ $$ \\ $$

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